Complex Numbers
If $z$ is a complex number, then the number of common roots of the equations $z^{1985}+z^{100}+1=0$ and $z^3+2 z^2+2 z+1=0$, is equal to
If $z=x+i y, x y \neq 0$, satisfies the equation $z^2+i \bar{z}=0$, then $\left|z^2\right|$ is equal to :
Let $\mathrm{r}$ and $\theta$ respectively be the modulus and amplitude of the complex number $z=2-i\left(2 \tan \frac{5 \pi}{8}\right)$, then $(\mathrm{r}, \theta)$ is equal to
If $z=\frac{1}{2}-2 i$ is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in \mathbb{R}$, then $\alpha+\beta$ is equal to
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 $(1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega$
$\begin{aligned} & A=1, B=1, C=0 \\ & \therefore 5(3 A-2 B-C)=5(3-2-0)=5 \end{aligned}$
the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to :
Let $S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$. Then $\sum_\limits{z \in \mathrm{S}}|z|^{2}$ is equal to :
Let $\mathrm{C}$ be the circle in the complex plane with centre $\mathrm{z}_{0}=\frac{1}{2}(1+3 i)$ and radius $r=1$. Let $\mathrm{z}_{1}=1+\mathrm{i}$ and the complex number $z_{2}$ be outside the circle $C$ such that $\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1$. If $z_{0}, z_{1}$ and $z_{2}$ are collinear, then the smaller value of $\left|z_{2}\right|^{2}$ is equal to :
For $a \in \mathbb{C}$, let $\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$ and $\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$. Then among the two statements :
(S1): If $\operatorname{Re}(a), \operatorname{Im}(a) > 0$, then the set A contains all the real numbers
(S2) : If $\operatorname{Re}(a), \operatorname{Im}(a) < 0$, then the set B contains all the real numbers,
Let $w_{1}$ be the point obtained by the rotation of $z_{1}=5+4 i$ about the origin through a right angle in the anticlockwise direction, and $w_{2}$ be the point obtained by the rotation of $z_{2}=3+5 i$ about the origin through a right angle in the clockwise direction. Then the principal argument of $w_{1}-w_{2}$ is equal to :
Let $S = \left\{ {z = x + iy:{{2z - 3i} \over {4z + 2i}}\,\mathrm{is\,a\,real\,number}} \right\}$. Then which of the following is NOT correct?
Let the complex number $z = x + iy$ be such that ${{2z - 3i} \over {2z + i}}$ is purely imaginary. If ${x} + {y^2} = 0$, then ${y^4} + {y^2} - y$ is equal to :
Let $A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$ is purely imaginary $\}$. Then the sum of the elements in $\mathrm{A}$ is :
If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation :
Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right.$ and $\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\}$ is equal to :
Let $a,b$ be two real numbers such that $ab < 0$. IF the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a+ib$ lies on the circle $|z-1|=|2z|$, then a possible value of $\frac{1+[a]}{4b}$, where $[t]$ is greatest integer function, is :
If the center and radius of the circle $\left| {{{z - 2} \over {z - 3}}} \right| = 2$ are respectively $(\alpha,\beta)$ and $\gamma$, then $3(\alpha+\beta+\gamma)$ is equal to :
For all $z \in C$ on the curve $C_{1}:|z|=4$, let the locus of the point $z+\frac{1}{z}$ be the curve $\mathrm{C}_{2}$. Then :
For two non-zero complex numbers $z_{1}$ and $z_{2}$, if $\operatorname{Re}\left(z_{1} z_{2}\right)=0$ and $\operatorname{Re}\left(z_{1}+z_{2}\right)=0$, then which of the following are possible?
A. $\operatorname{Im}\left(z_{1}\right)>0$ and $\operatorname{Im}\left(z_{2}\right) > 0$
B. $\operatorname{Im}\left(z_{1}\right) < 0$ and $\operatorname{Im}\left(z_{2}\right) > 0$
C. $\operatorname{Im}\left(z_{1}\right) > 0$ and $\operatorname{Im}\left(z_{2}\right) < 0$
D. $\operatorname{Im}\left(z_{1}\right) < 0$ and $\operatorname{Im}\left(z_{2}\right) < 0$
Choose the correct answer from the options given below :
Let $z$ be a complex number such that $\left| {{{z - 2i} \over {z + i}}} \right| = 2,z \ne - i$. Then $z$ lies on the circle of radius 2 and centre :
Let $\mathrm{z_1=2+3i}$ and $\mathrm{z_2=3+4i}$. The set $\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}}$ represents a
The value of ${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$ is
Let $\mathrm{p,q\in\mathbb{R}}$ and ${\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}(p + iq),i = \sqrt { - 1} $ then $\mathrm{p+q+q^2}$ and $\mathrm{p-q+q^2}$ are roots of the equation.
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
$B:{x^2} + {(y + 3)^2} = 16$
$A \cap B = (0, - 7)$
${\mathop{\rm Re}\nolimits} z - lm\,z = 7$
If $z \neq 0$ be a complex number such that $\left|z-\frac{1}{z}\right|=2$, then the maximum value of $|z|$ is :
Let $\mathrm{S}=\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z+i|=|z-1|\}$. Then the set of all values of $x$, for which $w=2 x+i y \in \mathrm{S}$ for some $y \in \mathbb{R}$, is :
If $z=2+3 i$, then $z^{5}+(\bar{z})^{5}$ is equal to :
Let $S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}$ and $S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}$. Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$, the least value of $\left|z_{2}-z_{1}\right|$ is :
Let S be the set of all $(\alpha, \beta), \pi<\alpha, \beta<2 \pi$, for which the complex number $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real. Let $Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$. Then $\sum\limits_{(\alpha, \beta) \in S}\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to :
Let the minimum value $v_{0}$ of $v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$ is attained at ${ }{z}=z_{0}$. Then $\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$ is equal to :
If $z=x+i y$ satisfies $|z|-2=0$ and $|z-i|-|z+5 i|=0$, then :
Let O be the origin and A be the point ${z_1} = 1 + 2i$. If B is the point ${z_2}$, ${\mathop{\rm Re}\nolimits} ({z_2}) < 0$, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?
For $z \in \mathbb{C}$ if the minimum value of $(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|)$ is $5 \sqrt{2}$, then a value Question: of $p$ is _____________.
For $\mathrm{n} \in \mathbf{N}$, let $\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}$ and $\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}$. Then the number of elements in the set $\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\}$ is :
The real part of the complex number ${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$ is equal to :
Let arg(z) represent the principal argument of the complex number z. Then, |z| = 3 and arg(z $-$ 1) $-$ arg(z + 1) = ${\pi \over 4}$ intersect :
Let $\alpha$ and $\beta$ be the roots of the equation x2 + (2i $-$ 1) = 0. Then, the value of |$\alpha$8 + $\beta$8| is equal to :
The number of points of intersection of
$|z - (4 + 3i)| = 2$ and $|z| + |z - 4| = 6$, z $\in$ C, is :
