${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$ is equal to :
($z,\,{z_1}\,\& \,{z_2}\,$ are complex numbers) will be :
Let the circle be $\left| {z - {z_3}} \right| = r.$
Let $z=(1+i)(1+2 i)(1+3 i) \ldots .(1+n i)$, where $i=\sqrt{-1}$. If $|z|^2=44200$, then $n$ is equal to $\_\_\_\_$
Explanation:
$ \begin{aligned} & Z=(1+i)(1+2 i)(1+3 i) \ldots(1+n i) \\ & |Z|=|(1+i)(1+2 i)(1+3 i) \ldots(1+n i)| \\ & |Z|=|1+i| \cdot|1+2 i| \cdot|1+3 i| \ldots|1+n i| \\ & |Z|^2=|1+i|^2 \cdot|1+2 i|^2 \cdot|1+3 i|^2 \ldots|1+n i|^2 \\ & |Z|^2=\left(1^2+1^2\right)\left(1^2+2^2\right)\left(1^2+3^2\right) \ldots\left(1^2+n^2\right)\left\{\text { as } z=x+i y,|z|^2=x^2+y^2\right\} \end{aligned} $
It is given that
$ \begin{aligned} & |Z|^2=44200 \\ & \left(1+1^2\right)\left(1+2^2\right)\left(1+3^2\right) \ldots\left(1+n^2\right)=44200 \\ & 44200=2^3 \cdot 5^2 \cdot 13 \cdot 17 \\ & \left(1+1^2\right)\left(1+2^2\right)\left(1+3^2\right) \cdot\left(1+4^2\right)\left(1+5^2\right) \ldots\left(1+n^2\right)=2^3 \cdot 5^2 \cdot 13 \cdot 17 \end{aligned} $
solve some product terms
$ 2 \cdot 5 \cdot(10) \cdot(17) \cdot(26) \ldots\left(1+n^2\right)=(2)(5)(2 \times 5)(17)(2 \times 13) $
this matches for $n=5$
so value of $n=5$
Let $\alpha=\frac{-1+i \sqrt{3}}{2}$ and $\beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$. If
$ (7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=m^{10}, $
then $m$ is $\_\_\_\_$
Explanation:
$ \begin{aligned} & Z=(7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7-9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20} \\ & =\alpha^{20}(7 \beta-7+9 \alpha)^{20}+\beta^{20}(-7+9 \alpha+7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+7^{20} \\ & =(-7+9 \alpha+7 \beta)^{20}(0)+7^{20} \\ & m^{10}=7^{20} \\ & m=49 \end{aligned} $
If $\alpha$ is a root of the equation $x^2+x+1=0$ and $\sum_\limits{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20$, then n is equal to _________.
Explanation:
$\begin{aligned} &\alpha \text { is root of equation } 1+x+x^2=0, \alpha=\omega \text { or } \omega^2\\ &\begin{aligned} & \left(\alpha^k+\frac{1}{\alpha^k}\right)^2=\alpha^{2 k}+\frac{1}{\alpha^{2 k}}+2=\omega^k+\frac{1}{\omega^k}+2 \\ & \Rightarrow \quad \omega^k+\frac{1}{\omega^k}+2=\left\{\begin{array}{l} 4,3 \text { divides } k \\ 1,3 \text { does not divide } k \end{array}\right. \\ & \therefore \quad \sum_{k=1}^n\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20 \\ & \Rightarrow \quad(1+1+4)+(1+1+4)+(1+1+4)+(1+1) \\ & =20 \\ & \Rightarrow \quad n=11 \end{aligned} \end{aligned}$
Let $\mathrm{A}=\{z \in \mathrm{C}:|z-2-i|=3\}, \mathrm{B}=\{z \in \mathrm{C}: \operatorname{Re}(z-i z)=2\}$ and $\mathrm{S}=\mathrm{A} \cap \mathrm{B}$. Then $\sum_{z \in S}|z|^2$ is equal to _________.
Explanation:
$\begin{aligned} &\text { Let } z=x+i y\\ &\begin{aligned} & |z-2-i|=3 \Rightarrow(x-2)^2+(y-1)^2=3^2 \\ & \operatorname{Re}(z-i z)=\operatorname{Re}(x+i y-i x+y)=x+y \Rightarrow x+y=2 \\ & \Rightarrow A=\left\{(x, y):(x-2)^2+(y-1)^2=3^2, x, y \in R\right\}, \\ & B=\{(x, y): x+y=2\} \\ & \Rightarrow x-2=-y \Rightarrow y^2+(y-1)^2=3^2 \\ & \Rightarrow 2 y^2-2 y-8=0 \Rightarrow y^2-y-4=0 \\ & y_1+y_2=1, y_1 y_2=-4 \\ & \Rightarrow y_1^2+y_2^2 \\ & =\left(y_1+y_2\right)^2-2 y_1 y_2=9 \\ & \Rightarrow x_1+x_2=4\left(y_1+y_2\right)=3, \\ & x_1 x_2=\left(2-y_1\right)\left(2-y_2\right)=4-2\left(y_1+y_2\right)+y_1 y_2=-2 \\ & \Rightarrow x_1^2+x_2^2=\left(x_1+x_2\right)^2-2 x_1 x_2=13 \\ & \because S=\left\{\left(x_1, y_1\right),\left(x_2, y_2\right)\right\} \\ & \Rightarrow \sum_{z \in S}|z|^2=\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)=22 \end{aligned} \end{aligned}$
Explanation:
$\begin{aligned} & a, b \in I,-3 \leq a, b \leq 3, a+b \neq 0 \\ & |z-a|=|z+b| \\ & \left|\begin{array}{ccc} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \\ & \Rightarrow\left|\begin{array}{ccc} z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \end{aligned}$
$\begin{aligned} & \Rightarrow z\left|\begin{array}{ccc} 1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \\ & \Rightarrow z\left|\begin{array}{ccc} 1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2 \end{array}\right|=1 \end{aligned}$
$\begin{aligned} &\begin{aligned} & \Rightarrow z^3=1 \\ & \Rightarrow z=\omega, \omega^2, 1 \end{aligned}\\ &\text { Now }\\ &\begin{aligned} & |1-\mathrm{a}|=|1+\mathrm{b}| \\ & \Rightarrow 10 \text { pairs } \end{aligned} \end{aligned}$
Let $\alpha, \beta$ be the roots of the equation $x^2-\mathrm{ax}-\mathrm{b}=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $\mathrm{P}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}$. If $\mathrm{P}_3=-5 \sqrt{7} i, \mathrm{P}_4=-3 \sqrt{7} i, \mathrm{P}_5=11 \sqrt{7} i$ and $\mathrm{P}_6=45 \sqrt{7} i$, then $\left|\alpha^4+\beta^4\right|$ is equal to __________.
Explanation:
We begin with the equations for the roots:
$\alpha + \beta = \mathrm{a}$
$\alpha \beta = -\mathrm{b}$
Given:
$ \mathrm{P}_6 = \mathrm{aP}_5 + \mathrm{bP}_4 $
$ \mathrm{P}_5 = \mathrm{aP}_4 + \mathrm{bP}_3 $
Using the given values:
For $\mathrm{P}_6$:
$ 45 \sqrt{7} i = \mathrm{a} \times 11 \sqrt{7} i + \mathrm{b}(-3 \sqrt{7}) i $
Simplifying, we obtain:
$ 45 = 11 \mathrm{a} - 3 \mathrm{b} \quad \text{(Equation 1)} $
For $\mathrm{P}_5$:
$ 11 \sqrt{7} i = \mathrm{a}(-3 \sqrt{7} i) + \mathrm{b}(-5 \sqrt{7} i) $
Simplifying, we obtain:
$ 11 = -3 \mathrm{a} - 5 \mathrm{b} \quad \text{(Equation 2)} $
Solving these linear equations, we find:
$ \mathrm{a} = 3 $
$ \mathrm{b} = -4 $
Now, we calculate $\left|\alpha^4 + \beta^4\right|$ using the relation:
$ \left|\alpha^4 + \beta^4\right| = \sqrt{(\alpha^4 - \beta^4)^2 + 4 (\alpha^4 \beta^4)} $
From $\mathrm{b} = -4$, we know:
$ \alpha \beta = -\mathrm{b} = 4 \quad \Rightarrow \alpha^4 \beta^4 = (\alpha \beta)^4 = 4^4 = 256 $
Substitute into the relation:
$ \left|\alpha^4 + \beta^4\right| = \sqrt{(-63) + 1024} $
$ = \sqrt{961} = 31 $
Thus, $\left|\alpha^4 + \beta^4\right|$ is equal to 31.
The sum of the square of the modulus of the elements in the set $\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathbf{Z}, z \in \mathbf{C},|z-1| \leq 1,|z-5| \leq|z-5 \mathrm{i}|\}$ is __________.
Explanation:
$z$ should be lying in the shaded region shown in adjacent figure.
Possible $z$ are
$z=0+0 i,(1+0 i),(2+0 i),(1-i),(1+i)$
Sum of squares of modulus
$\begin{aligned} & =0+1+4+2+2 \\ & =9 \end{aligned}$
Explanation:
Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}_1\right)$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$
Circle : $(x+2)^2+(y-3)^2=1$
$ \begin{aligned} & \mathrm{L}_1: \mathrm{x}+\mathrm{y}-1=0 \\\\ & \mathrm{~L}_2: \mathrm{x}-\mathrm{y}+4=0 \end{aligned} $
On solving circle $\& \mathrm{~L}_1$ we get
$ \mathrm{z}_1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right) $
On solving $\mathrm{L}_1$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$ we get $z_2:\left(\frac{-3}{2}, \frac{5}{2}\right)$
$\begin{aligned} & \left|z_1\right|^2+2\left|z_2\right|^2=14+5 \sqrt{2}+17 \\\\ & =31+5 \sqrt{2}\end{aligned}$
$\begin{array}{ll}\text { So } & \alpha=31 \\\\ & \beta=5 \\\\ & \alpha+\beta=36\end{array}$
If $\alpha$ denotes the number of solutions of $|1-i|^x=2^x$ and $\beta=\left(\frac{|z|}{\arg (z)}\right)$, where $z=\frac{\pi}{4}(1+i)^4\left[\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right], i=\sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4 x-3 y=7$ is __________.
Explanation:
$\begin{aligned} & (\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1 \\ & z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right] \\ & =-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right) \\ & =2 \pi i \\ & \beta=\frac{2 \pi}{\frac{\pi}{2}}=4 \end{aligned}$
Distance from $(1,4)$ to $4 x-3 y=7$
Will be $\frac{15}{5}=3$
Let $\alpha, \beta$ be the roots of the equation $x^2-\sqrt{6} x+3=0$ such that $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+i b), i=\sqrt{-1}$. Then $n+a+b$ is equal to __________.
Explanation:
$\begin{aligned} & x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \\ & \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \\ & \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) \\ & =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} \\ & =3^{49}(-1+i) \\ & =3^n(a+i b) \\ & \therefore n=49, a=-1, b=1 \\ & \therefore n+a+b=49-1+1=49 \end{aligned}$
Let $\alpha, \beta$ be the roots of the equation $x^2-x+2=0$ with $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Then $\alpha^6+\alpha^4+\beta^4-5 \alpha^2$ is equal to ___________.
Explanation:
$\begin{aligned} & \alpha^6+\alpha^4+\beta^4-5 \alpha^2 \\ & =\alpha^4(\alpha-2)+\alpha^4-5 \alpha^2+(\beta-2)^2 \\ & =\alpha^5-\alpha^4-5 \alpha^2+\beta^2-4 \beta+4 \\ & =\alpha^3(\alpha-2)-\alpha^4-5 \alpha^2+\beta-2-4 \beta+4 \\ & =-2 \alpha^3-5 \alpha^2-3 \beta+2 \\ & =-2 \alpha(\alpha-2)-5 \alpha^2-3 \beta+2 \\ & =-7 \alpha^2+4 \alpha-3 \beta+2 \\ & =-7(\alpha-2)+4 \alpha-3 \beta+2 \\ & =-3 \alpha-3 \beta+16=-3(1)+16=13 \end{aligned}$
Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $\left|z-z_0\right|^2=4$ and $\left|z-z_0\right|^2=16$ respectively, where $z_0=1+i$. Then, the value of $100|\alpha|^2$ is __________.
Explanation:
$\begin{aligned} & \left|z-z_0\right|^2=4 \\ & \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 \\ & \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 \\ & \Rightarrow|\alpha|^2-\alpha \bar{z}_0-z_0 \bar{\alpha}=2 \quad\text{......... (1)} \\ & \left|z-z_0\right|^2=16 \end{aligned}$
$\begin{aligned} & \Rightarrow\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=16 \\ & \Rightarrow\left(1-\bar{\alpha} z_0\right)\left(1-\alpha \bar{z}_0\right)=16|\alpha|^2 \\ & \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0+|\alpha|^2\left|z_0\right|^2=16|\alpha|^2 \\ & \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0=14|\alpha|^2 \quad\text{....... (2)} \end{aligned}$
From (1) and (2)
$\begin{aligned} & \Rightarrow 5|\alpha|^2=1 \\ & \Rightarrow 100|\alpha|^2=20 \end{aligned}$
Explanation:
$x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha$
Let $\alpha=\omega$
Now $(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$
$\begin{aligned} & A=1, B=1, C=0 \\ & \therefore 5(3 A-2 B-C)=5(3-2-0)=5 \end{aligned}$
Let $w=z \bar{z}+k_{1} z+k_{2} i z+\lambda(1+i), k_{1}, k_{2} \in \mathbb{R}$. Let $\operatorname{Re}(w)=0$ be the circle $\mathrm{C}$ of radius 1 in the first quadrant touching the line $y=1$ and the $y$-axis. If the curve $\operatorname{Im}(w)=0$ intersects $\mathrm{C}$ at $\mathrm{A}$ and $\mathrm{B}$, then $30(A B)^{2}$ is equal to __________
Explanation:
$w = z\bar{z} + k_1z + k_2iz + \lambda(1+i), \quad \text{where } k_1, k_2 \in \mathbb{R}.$
1. If $w = x+iy$, we can separate this into the real and imaginary parts :
The real part is: $\text{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0.$
The imaginary part is: $\text{Im}(w) = k_1y + k_2x + \lambda = 0.$
2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$.
3. By comparing this with $\text{Re}(w) = 0$, we can see that $k_1=-2$, $k_2=4$, and $\lambda=4$.
4. The equation for the curve $\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$, which simplifies to $y = 2x + 2$.
5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$, with solutions $x = 0$ and $x = 2/5$.
6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$.
7. The distance between A and B is $AB = \sqrt{[(2/5)^2 + (4/5)^2]} = \sqrt{4/5}$.
8. Finally, $30(AB)^2 = 30 \times (4/5) = 24$.
Let $\mathrm{S}=\left\{z \in \mathbb{C}-\{i, 2 i\}: \frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \in \mathbb{R}\right\}$. If $\alpha-\frac{13}{11} i \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\}$, then $242 \alpha^{2}$ is equal to _________.
Explanation:
$ \begin{aligned} & \operatorname{lm}\left(\frac{z^2+8 i z-15}{z^2-3 i z-2}\right)=0 \\\\ & \Rightarrow-\left(x^2-y^2-8 y-15\right)(2 x y-3 x)+(2 x y+8 x)\left(x^2-\right. \left.y^2+3 y-2\right)=0 \\\\ & \Rightarrow\left(x^2-y^2\right)(2 x y+8 x-2 x y+3 x)+(8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\ & \Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y-45 x+6 x y^2 -4 x y+24 x y-16 x=0 \\\\ & \Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\ & \Rightarrow\left(11 x^2+11 y^2+26 y-61)=0\right. \\\\ & \because \alpha \neq 0, \Rightarrow x=0 \text { (neglected) } \\\\ & \text { Put } y=-\frac{13}{11}, \quad x=\alpha \\\\ & 11 \alpha^2+11 \cdot \frac{13^2}{11^2}-26 \cdot \frac{13}{11}-61=0 \\\\ & \Rightarrow 121 \alpha^2=840 \\\\ & \Rightarrow 242 \alpha^2=1680 \end{aligned} $
For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$, if $\sqrt{\lambda-1}$ is the radius of the circle $|z-\alpha|^{2}+|z-\beta|^{2}=2 \lambda$, then $|\alpha-\beta|$ is equal to __________.
Explanation:
$ \begin{aligned} & \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\ & \therefore 2 \lambda=|\alpha-\beta|^2 .........(i) \end{aligned} $
For circle,
$ \left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2 $
$\begin{array}{lll}\text { Radius, } r=\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|=2 \sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|^2=4(\lambda-1) \\\\ \Rightarrow 2 \lambda=4(\lambda-1) [\because \text { Using Eq. (i)] } \\\\ \Rightarrow 2 \lambda=4 \\\\ \Rightarrow \lambda=2 \end{array}$
$\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}$
Let $z=1+i$ and $z_{1}=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$. Then $\frac{12}{\pi} \arg \left(z_{1}\right)$ is equal to __________.
Explanation:
$z = 1 + i$
${z_1} = {{1 + i\overline z } \over {\overline z (1 - z) + {1 \over z}}}$
$ = {{z(1 + i\overline z )} \over {|z{|^2}(1 - z) + 1}}$
$ = {{(1 + i)(1 + i(1 - i))} \over {2(1 - 1 - i) + 1}}$
${z_1} = 1 - i$
$\arg {z_1} = {\tan ^{ - 1}}\left( {{{ - 1} \over 1}} \right) = {{3\pi } \over 4}$
${{12} \over \pi }\arg ({z_1}) = {{3\pi } \over 4}\,.\,{{12} \over \pi } = 9$
Let $\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}$ and $B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}$. Then $\sum\limits_{z \in A \cap B} {({\mathop{\rm Re}\nolimits} z - {\mathop{\rm Im}\nolimits} z)} $ is equal to ____________.
Explanation:
Let $z = x + iy$
and $\alpha = 8 - 14i$
${{\alpha z - \overline \alpha \,\overline z } \over {{z^2} - {{\overline z }^2} - 112i}} = 1$
$\therefore$ ${{(16y - 28x)} \over {4xy - 112i}} = 1$
$(16y - 28x + 112)i = 4xy$
$\therefore$ $z = - 7i$ or 4
Now, $z = - 7i$ satisfy B
$B:{x^2} + {(y + 3)^2} = 16$
$A \cap B = (0, - 7)$
${\mathop{\rm Re}\nolimits} z - lm\,z = 7$
Let $\mathrm{z}=a+i b, b \neq 0$ be complex numbers satisfying $z^{2}=\bar{z} \cdot 2^{1-z}$. Then the least value of $n \in N$, such that $z^{n}=(z+1)^{n}$, is equal to __________.
Explanation:
$\because$ ${z^2} = \overline z \,.\,{2^{1 - |z|}}$ ...... (1)
$ \Rightarrow |z{|^2} = |\overline z |\,.\,{2^{1 - |z|}}$
$ \Rightarrow |z| = {2^{1 - |z|}}$,
$\because$ $b \ne 0 \Rightarrow |z| \ne 0$
$\therefore$ $|z| = 1$ ...... (2)
$\because$ $z = a + ib$ then $\sqrt {{a^2} + {b^2}} = 1$ ...... (3)
Now again from equation (1), equation (2), equation (3) we get :
${a^2} - {b^2} + i2ab = (a - ib){2^0}$
$\therefore$ ${a^2} - {b^2} = a$ and $2ab = - b$
$\therefore$ $a = - {1 \over 2}$ and $b = \, \pm \,{{\sqrt 3 } \over 2}$
$z = - {1 \over 2} + {{\sqrt 3 } \over 2}i$ or $z = - {1 \over 2} - {{\sqrt 3 } \over 2}i$
${z^n} = {(z + 1)^n} \Rightarrow {\left( {{{z + 1} \over z}} \right)^n} = 1$
${\left( {1 + {1 \over z}} \right)^n} = 1$
$\left( {{{1 + \sqrt 3 i} \over 2}} \right) = 1$, then minimum value of n is 6.
Let $S=\left\{z \in \mathbb{C}: z^{2}+\bar{z}=0\right\}$. Then $\sum\limits_{z \in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$ is equal to ______________.
Explanation:
$\because$ ${z^2} + \overline z = 0$
Let $z = x + iy$
$\therefore$ ${x^2} - {y^2} + 2ixy + x - iy = 0$
$({x^2} - {y^2} + x) + i(2xy - y) = 0$
$\therefore$ ${x^2} + {y^2} = 0$ and $(2x - 1)y = 0$
if $x = \, + \,{1 \over 2}$ then $y = \, \pm \,{{\sqrt 3 } \over 2}$
And if $y = 0$ then $x = 0, - 1$
$\therefore$ $z = 0 + 0i, - 1 + 0i,{1 \over 2} + {{\sqrt 3 } \over 2}i,{1 \over 2} - {{\sqrt 3 } \over 2}i$
$\therefore$ $\sum {\left( {{R_e}(z) + m(z)} \right) = 0} $
Let $S = \{ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\} $. Let $|z - 4i|$ attains minimum and maximum values, respectively, at z1 $\in$ S and z2 $\in$ S. If $5(|{z_1}{|^2} + |{z_2}{|^2}) = \alpha + \beta \sqrt 5 $, where $\alpha$ and $\beta$ are integers, then the value of $\alpha$ + $\beta$ is equal to ___________.
Explanation:
$S$ represents the shaded region shown in the diagram.
Clearly $z_{1}$ will be the point of intersection of $P A$ and given circle.
$P A: 2 x+y=4$ and given circle has equation $(x-2)^{2}+y^{2}=1$
On solving we get
$z_{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2}{\sqrt{5}} i \Rightarrow\left|z_{1}\right|^{2}=5-\frac{4}{\sqrt{5}}$
$z_{2}$ will be either $B$ or $C$.
$\because P B=\sqrt{17}$ and $P C=\sqrt{13}$ hence $z_{2}=1$
So $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=30-4 \sqrt{5}$
Clearly $\alpha=30$ and $\beta=-4 \Rightarrow \alpha+\beta=26$
Sum of squares of modulus of all the complex numbers z satisfying $\overline z = i{z^2} + {z^2} - z$ is equal to ___________.
Explanation:
So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and
$ x^{2}-y^{2}+2 x y=0\quad\dots(ii) $
From (i) and (ii) we get
$ x=0 \text { or } y=-\frac{1}{2} $
When $x=0$ we get $y=0$
When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$
So there will be total 3 possible values of $z$, which are $0,\left(\frac{-1+\sqrt{2}}{2}\right)-\frac{1}{2} i$ and $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$
Sum of squares of modulus
$ \begin{aligned} &=0+\left(\frac{\sqrt{2}-1}{2}\right)^{2}+\frac{1}{4}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=+\frac{1}{4} \\\\ &=2 \end{aligned} $
The number of elements in the set {z = a + ib $\in$ C : a, b $\in$ Z and 1 < | z $-$ 3 + 2i | < 4} is __________.
Explanation:
at line $y=-2$, we have $(5,-2)(6,-2)(1,-2)(0,-2)$ $\Rightarrow 4$ points
at line $y=-1$, we have $(4,-1)(5,-1)(6,-1)(2,-1)$ $(1,-1)(0,-1) \Rightarrow 6$ points
at line $y=0$, we have $(0,0)(1,0)(2,0)(3,0)(4,0)$ $(5,0)(6,0) \Rightarrow 7$ points
at line $y=1$, we have $(1,1),(2,1),(3,1),(4,1),(5,1)$ i.e. 5 points
symmetrically
at line $y=-5$, we have 5 points
at line $y=-4$, we have 7 points
at line $y=-3$, we have 6 points
So Total integral points $=2(5+7+6)+4$
$ =40 $
If ${z^2} + z + 1 = 0$, $z \in C$, then
$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$ is equal to _________.
Explanation:
$\because$ ${z^2} + z + 1 = 0$
$\Rightarrow$ $\omega$ or $\omega$2
$\because$ $\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$
$ = \left| {\sum\limits_{n = 1}^{15} {{z^{2n}} + \sum\limits_{n = 1}^{15} {{z^{ - 2n}} + 2\,.\,\sum\limits_{n = 1}^{15} {{{( - 1)}^n}} } } } \right|$
$ = \left| {0 + 0 - 2} \right|$
$ = 2$
Let S = {z $\in$ C : |z $-$ 3| $\le$ 1 and z(4 + 3i) + $\overline z $(4 $-$ 3i) $\le$ 24}. If $\alpha$ + i$\beta$ is the point in S which is closest to 4i, then 25($\alpha$ + $\beta$) is equal to ___________.
Explanation:
Here $|z - 3| < 1$
$ \Rightarrow {(x - 3)^2} + {y^2} < 1$
and $z = (4 + 3i) + \overline z (4 - 3i) \le 24$
$ \Rightarrow 4x - 3y \le 12$
$\tan \theta = {4 \over 3}$
$\therefore$ Coordinate of $P = (3 - \cos \theta ,\sin \theta )$
$ = \left( {3 - {3 \over 5},{4 \over 5}} \right)$
$\therefore$ $\alpha + i\beta = {{12} \over 5} + {4 \over 5}i$
$\therefore$ $25(\alpha + \beta ) = 80$
Explanation:
| x + iy $-$ 2 $-$ 2i | $\le$ 1
|(x $-$ 2) + i(y $-$ 2)| $\le$ 1
(x $-$ 2)2 + (y $-$ 2)2 $\le$ 1
| 3iz + 6 |max at a + ib
$\left| {3i} \right|\left| {z + {6 \over {3i}}} \right|$
$3{\left| {z - 2i} \right|_{\max }}$
From figure maximum distance at 3 + 2i
a + ib = 3 + 2i = a + b = 3 + 2 = 5 Ans.
Explanation:
$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$
$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$
${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$
${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$
${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$
$4y = {x^2} - 4 + {y^2}$
${x^2} + {y^2} - 4y - 4 = 0$
locus is a circle with center (0, 2) & radius = $2\sqrt 2 $
min. value = ${(AP)^2} = {(OP - OA)^2}$
$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$
$ = {\left( {7\sqrt 2 } \right)^2} = 98$
Explanation:
z1 $-$ z2 = (x1 $-$ x2) + i(y1 $-$ y2)
$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$
${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)
$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$ .... (2)
$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$ .... (3)
sub (2) & (3)
${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$
$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$
$ = ({x_1} - {x_2})({x_1} + {x_2})$
${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$
Explanation:
$ = {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$
This is positive integer for n = 6
Explanation:
${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$
$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $
$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$
$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $
$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $
$ = 21 - 2 - 6$
$ = 13$
Explanation:
$\Rightarrow$ $\theta$ = ${{\pi \over 4}}$
Hence, sin23$\theta$ + cos2$\theta$ = 1.
Explanation:
x2 + y2 + 2x = 0
Centre : ($-$1, 0)
Parabola : x2 $-$ 6x $-$ y + 13 = 0
(x $-$ 3)2 = y $-$ 4
Vertex : (3, 4)
Equation of line $ \equiv y - 0 = {{4 - 0} \over {3 + 1}}(x + 1)$
$y = x + 1$
y-intercept = 1
Explanation:
$\Rightarrow$ AX = IX
$\Rightarrow$ A = I
$ \Rightarrow {\left( {\matrix{ 0 & i \cr 1 & 0 \cr } } \right)^n} = I$
$ \Rightarrow {A^8} = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$
$\Rightarrow$ n is multiple of 8
So, number of 2 digit numbers in the set
S = 11 (16, 24, 32, .........., 96)
Explanation:
$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$
Here one vertex z3 is 0
$ \therefore $ $z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$
Given, z1, z2 are roots of ${z^2} + az + 12 = 0$
$ \therefore $ ${z_1} + {z_2} = - a$
${z_1}{z_2} = 12$
$ \therefore $ $z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$
$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$
$ \Rightarrow {( - a)^2} = 3 \times 12$
$ \Rightarrow {a^2} = 36$
$ \Rightarrow a = \pm 6$
$ \Rightarrow |a|\, = 6$
Explanation:
| z + i | = | z $-$ 3i |
$ \Rightarrow $ y = 1
Now
$\omega$ = x2 + y2 $-$ 2x $-$ 2iy + 2
$\omega$ = x2 + 1 $-$ 2x $-$ 2i + 2
Re($\omega$) = x2 $-$ 2x + 3
Re($\omega$) = (x $-$ 1)2 + 2
Re($\omega$)min at x = 1 $ \Rightarrow $ z = 1 + i
Now,
$\omega$ = 1 + 1 $-$ 2 $-$ 2i + 2
$\omega$ = 2(1 $-$ i) = 2$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$
$\omega$n = 2$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$
If $\omega$n is real $ \Rightarrow $ n = 4
| z + 5 | $ \le $ 4 and z(1 + i) + $\overline z $(1 $-$ i) $ \ge $ $-$10, i = $\sqrt { - 1} $.
If the maximum value of | z + 1 |2 is $\alpha$ + $\beta$$\sqrt 2 $, then the value of ($\alpha$ + $\beta$) is ____________.
Explanation:
Given, z(1 + i) + $\overline z $ (1 $-$ i) $ \ge $ $-$ 10
$ \Rightarrow $ z + $\overline z $ + i (z $-$ $\overline z $) $ \ge $ $-$ 10
$ \Rightarrow $ 2x + i (2iy) $ \ge $ $-$ 10
$ \Rightarrow $ x + i2 y $ \ge $ $-$ 5
$ \Rightarrow $ x $-$ y $ \ge $ $-$ 5 ...... (1)
Also given, | z + 5 | $ \le $ 4
$ \Rightarrow $ | z $-$ ($-$5 + 0i) | $ \le $ 4 ...... (2)
It represents a circle whose center at ($-$ 5, 0) and radius 4. z is inside of the circle.
From (1) and (2) z is the shaded region of the diagram.
Now, | z + 1 | = | z $-$ ($-$1 + 0 i) | = distance of z from ($-$1, 0).
Clearly 'p' is the required position of 'z' when | z + 1 | is maximum.
$ \therefore $ P $ \equiv $ ($-$5 $-$ 4 cos45$^\circ$, 0 $-$ 4sin45$^\circ$) = ($-$5$-$2$\sqrt 2 $, $-$2$\sqrt 2 $)
$ \therefore $ (PQ)2|max = 32 + 16$\sqrt 2 $
$ \Rightarrow $ $\alpha$ = 32
$ \Rightarrow $ $\beta$ = 16
Thus, $\alpha$ + $\beta$ = 48
Explanation:
${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$
We know,
$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $
$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $
and $ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$
$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$
$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$
Now, $K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$
$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$
$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$ [as ${\omega ^3} = 1$, ${i^4} = 1$]
$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$
$ \therefore $ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$
Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $
= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $
= $\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $
= ${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$
= 55 + 135 + 120
= 310
equation z + $\alpha $|z – 1| + 2i = 0 (z $ \in $ C and i = $\sqrt { - 1} $) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.
Explanation:
$ \therefore $ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$
y = $-$2 & ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$
${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$
$x \in R \Rightarrow D \ge 0$
$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$
${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$
${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$
${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$
$0 \le {\alpha ^2} \le {5 \over 4}$
$ \therefore $ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$
$ \therefore $ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$
then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$
Explanation:
$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$
$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$ = i4 [ As i4 = 1]
$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$
$ \Rightarrow $ m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
$ \therefore $ GCD (8, 12) = 4
Match each entry in List-I to the correct entries in List-II.
| List - I | List - II |
|---|---|
| (P) $|z|^2$ is equal to | (1) 12 |
| (Q) $|z-\bar{z}|^2$ is equal to | (2) 4 |
| (R) $|z|^2+|z+\bar{z}|^2$ is equal to | (3) 8 |
| (S) $|z+1|^2$ is equal to | (4) 10 |
| (5) 7 |
The correct option is: