Complex Numbers

Let $\alpha$ be a purely imaginary root of $a z^2+z+1=0$

Then,   $ a \alpha^2+\alpha+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Taking conjugate on both sides, we get

$ \begin{array}{ll} & \bar{a}(\bar{\alpha})^2+\bar{\alpha}+1=0 \\ \Rightarrow & \bar{a} \alpha^2-\alpha+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \,\,\,\,\,\,\,\,\,[\because \bar{\alpha}=-\alpha]\end{array} $

Solving, Eqs. (i) and (ii) by cross-multiplication, we get

$ \begin{array}{rlrl} \frac{\alpha^2}{1+1} & =\frac{\alpha}{\bar{a}-a}=\frac{1}{-a-\bar{a}} \\ \Rightarrow & \alpha^2 =\frac{-2}{a+\bar{a}} \text { and } \alpha=\frac{\bar{a}-a}{-a-\bar{a}} \end{array} $

Now, $\alpha^2=(\alpha)^2$

$ \begin{aligned} & \Rightarrow \frac{-2}{a+\bar{a}}=\left(\frac{\bar{a}-a}{-a-\bar{a}}\right)^2=\frac{(\bar{a}-a)^2}{(a+\bar{a})^2} \\ & \Rightarrow-2(a+\bar{a})=(\bar{a}-a)^2 \\ & \Rightarrow(\bar{a}-a)^2+2(\bar{a}+a)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Now, $|a|=1 \Rightarrow a=\cos \theta+i \sin \theta$

Put the value of $a$ in Eq. (iii), we get

$ \begin{aligned} & (-2 i \sin \theta)^2+4 \cos \theta=0 \\ \Rightarrow & -4 \sin ^2 \theta+4 \cos \theta=0 \\ \Rightarrow & \cos ^2 \theta+\cos \theta-1=0 \\ \Rightarrow & \cos \theta=\frac{\sqrt{5}-1}{2}, \frac{-\sqrt{5}-1}{2} \quad[\because \cos \theta \in[-1,1]] \\ \Rightarrow & \quad \theta=\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right) \end{aligned} $

Hence, $a=\cos \left\{\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right\}$$ +i \sin \left\{\cos ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right\} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} We\,\,\, have, & \left|2 z \cos \theta+z^2\right| < |2 z \cos \theta|+\left|z^2\right| \\ & \quad\left(\because\left|z_1+z_2\right| \leq\left|z_1\right|+\left|z_2\right|\right) \\ & < 2|z||\cos \theta|+|z|^2 \\ & < 2|z|+|z|^2 \\ & < 2 k+k^2 \quad \quad[\because|z| < k] \end{aligned} \end{aligned} $

$ \begin{aligned} \text { But } & 2 z \cos \theta+z^2 \mid >1 \\ & \therefore \,\,2 k+k^2 >1 \\ \Rightarrow &\,\, k^2+2 k-1 >0 \\ \Rightarrow \,\,& k >\sqrt{2}-1 \end{aligned} $

According to question,

Let find slope of $ A B, B C $ and $ C A $

Slope of $ A B=\frac{1-3}{1-(-2)}=\frac{-2}{3} $

Slope of $ B C=\frac{3-\left(\frac{5}{3}\right)}{-2-0}=\frac{-2}{3} $

Slope of $ C A=\frac{\frac{5}{3}-1}{0-1}=\frac{-2}{3} $

Since, slope of all lines are same, therefore, points are collinear.

Let $ z=x+i y $

Given,

$ |z+3-i|=1 $

$ \begin{array}{lr} \Rightarrow & |(x+3)+(y-1) i|=1 \\ \Rightarrow & \sqrt{(x+3)^{2}+(y-1)^{2}}=1 \\ \Rightarrow & (x+3)^{2}+(y-1)^{2}=1 \end{array} \quad \ldots \text { (i) } $

Above is the equation of circle with centre $ \equiv(-3,1) $ and

radius $ =1 $.

Also, $ \arg (z)=\pi \Rightarrow \tan ^{-1}\left|\frac{y}{x}\right|=\pi $

$ \Rightarrow \quad\left|\frac{y}{x}\right|=\tan \pi=0 \Rightarrow y=0 $

By Eq. (i),

$ \begin{array}{lrl} \Rightarrow & x & =-3 \\ \therefore & z & =-3+0 i \\ \Rightarrow & \text { Modulus of } z & =3 \end{array} $

Given, $z^2+|z|^2=0, z \neq 0$

Let $z=x+i y$ and $\bar{z}=x-i y$

$\begin{array}{ll} \Rightarrow(x+i y)^2+(x+i y)(x-i y)=0 & \left\{\because z \bar{z}=|z|^2\right\} \\ \Rightarrow x^2+i^2 y^2+i 2 x y+x^2-i^2 y^2=0 \\ \Rightarrow x^2-y^2+i 2 x y+x^2+y^2=0 \quad \{\because i^2=-1\}\\ \Rightarrow 2 x(x+i y)=0 \\ \Rightarrow x=0 \text { or } x+i y=0 \text { (not possible) } \end{array}$

Therefore, $x=0$ and $y \neq 0$

So, $y$ can take any non-zero real value.

Hence, infinitely many solutions exists.

$\begin{aligned} & \text { Let } z=(1)^{1 / n}=(\cos 2 k \pi+i \sin 2 k \pi)^{1 / n} \\ & \Rightarrow \quad z=\left(\frac{2 k \pi}{n}\right)+i \sin \left(\frac{2 k \pi}{n}\right), k=0,1,2 \ldots, n-1 \\ & \text { Let } z_1=\cos \left(\frac{2 k_1 \pi}{n}\right)+i \sin \left(\frac{2 k_1 \pi}{n}\right) \\ & \text { and } z_2=\cos \left(\frac{2 k_2 \pi}{n}\right)+i \sin \left(\frac{2 k_2 \pi}{n}\right) \end{aligned}$

be the two value of $z$ such that they subtend right angle at origin.

$\therefore \frac{2 k_1 \pi}{n}-\frac{2 k_2 \pi}{n}= \pm \frac{\pi}{2} \Rightarrow 4\left(k_1-k_2\right)= \pm n$

As, $k_1$ and $k_2$ are integer and $k_1 \neq k_2$

$\therefore n=4 k, k \in I$

To solve this, we need to first express $w$ in the polar form since we know the modulus (or absolute value) of $w$.

$w = |w|e^{iθ} = 2e^{iθ}$

where $θ$ is the argument of $w$.

Substitute this into the expression for $z$:

$z = w - {1 \over w} = 2e^{iθ} - {1 \over 2e^{iθ}}$

By applying Euler's formula, which states that $e^{iθ} = cos(θ) + isin(θ)$, the expression becomes :

$z = 2[cos(θ) + isin(θ)] - {1 \over 2}[cos(θ) - isin(θ)]$

Simplify the expression :

$z = [2cos(θ) - {1 \over 2}cos(θ)] + i[2sin(θ) + {1 \over 2}sin(θ)]$

$z = {3 \over 2}cos(θ) + {5 \over 2}isin(θ)$

The modulus of $z$ can be found using the formula

$|z| = \sqrt{(Re(z))^2 + (Im(z))^2}$, where $Re(z)$ and $Im(z)$ are the real and imaginary parts of $z$, respectively.

So,

$|z| = \sqrt{(\frac{3}{2} \cos(θ))^2 + (\frac{5}{2} \sin(θ))^2} = \sqrt{(\frac{9}{4} \cos^2(θ)) + (\frac{25}{4} \sin^2(θ))}$

We know that $\sin^2(θ) + \cos^2(θ) = 1$ (based on the Pythagorean trigonometric identity), so let's replace $\cos^2(θ)$ with $1 - \sin^2(θ)$ in the equation above.

$|z| = \sqrt{(\frac{9}{4} (1 - \sin^2(θ))) + (\frac{25}{4} \sin^2(θ))}$

$= \sqrt{(\frac{9}{4} - \frac{9}{4} \sin^2(θ)) + (\frac{25}{4} \sin^2(θ))}$

$= \sqrt{\frac{9}{4} + \frac{16}{4} \sin^2(θ)}$

$= \sqrt{\frac{9 + 16 \sin^2(θ)}{4}}$

$ = \frac{1}{2} \sqrt{9 + 16 \sin^2(θ)}$

Now, $\sin^2(θ)$ is always less than or equal to 1, so $16 \sin^2(θ)$ is always less than or equal to 16.

So, the maximum value of $|z|$ is when $\sin^2(θ) = 1$, which makes $|z|$ equal to $\frac{1}{2} \sqrt{9 + 16} = \frac{1}{2} \sqrt{25} = \frac{5}{2}$.

So the set of points $z$ satisfying $|z| ≤ 3$ (Option D) includes all points $z$, and thus it is the correct choice.

${\left[ {{{1 + \sin {\pi \over 8} + i\cos {\pi \over 8}} \over {1 + \sin {\pi \over 8} - i\cos {\pi \over 8}}}} \right]^n} = {\left[ {{{1 + \cos \alpha + i\sin \alpha } \over {1 + \cos \alpha - i\sin \alpha }}} \right]^n}$ [$\because$ $\alpha = {\pi \over 2} - {\pi \over 8}$]

$ = {\left[ {{{2{{\cos }^2}{\alpha \over 2} + 2i\sin {\alpha \over 2}\cos {\alpha \over 2}} \over {2{{\cos }^2}{\alpha \over 2} - 2i\sin {\alpha \over 2}\cos {\alpha \over 2}}}} \right]^n}$

$ = \left[ {{{\cos {\alpha \over 2} + i\sin {\alpha \over 2}} \over {\cos {\alpha \over 2} - i\sin {\alpha \over 2}}}} \right]_ - ^n = {\left( {{e^{2t{\alpha \over 2}}}} \right)^n} = {e^{in\,\alpha }}$

$ = {e^{in\left( {{{3\pi } \over 8}} \right)}} = \cos {{3n\pi } \over 8} + i\sin {{3n\pi } \over 8}$

For n = 4 we get imaginary part.

Let z = x + iy

Re(z + 2) = | z $-$ 2 |

Re(x + iy + 2) = | x + iy $-$ 2 |

x + 2 = $\sqrt {{{(x - 2)}^2} + {y^2}} $

Squaring on both sides,

(x + 2)² = (x $-$ 2)² + y²

x² + 4 + 4x = x² + 4 $-$ 4x + y² $\Rightarrow$ y² = 8x

The locus of z is a parabola.

We have, $Z = {{7 + i} \over {3 + 4i}}$

$ \Rightarrow Z = \left( {{{7 + i} \over {3 + 4i}}} \right)\left( {{{3 - 4i} \over {3 - 4i}}} \right) = {{21 + 4 - 28i + 3i} \over {9 + 16}} = 1 - i$

$\therefore$ ${Z^{14}} = {(1 - i)^{14}} = {[{(1 - i)^2}]^7} = {(1 - 1 - 2i)^7}$

$ = {( - 2)^7}{(i)^7} = {2^7}i$

The given equation is

$2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0$

$ \Rightarrow x = {{4(2 - i) \pm \sqrt {16{{(2 - i)}^2} + 8(1 + i)(5 + 3i)} } \over {4(1 + i)}}$

$ = - {i \over {1 + i}}$ or ${{4 - i} \over {1 + i}} = {{ - 1 - i} \over 2}$ or, ${{3 - 5i} \over 2}$

Now, $\left| {{{ - 1 - i} \over 1}} \right| = \sqrt {{1 \over 4} + {1 \over 4}} = \sqrt {{1 \over 2}} $

and $\left| {{{3 - 5i} \over 2}} \right| = \sqrt {{9 \over 4} + {{25} \over 4}} = \sqrt {{{17} \over 2}} $

Also, $\sqrt {{{17} \over 2}} > \sqrt {{1 \over 2}} $

Hence, required root is ${{3 - 5i} \over 2}$.

We have,

$Z = r{e^{i\theta }}$

$Z = r(\cos \theta + i\sin \theta )$

${e^{iz}} = {e^{ir(\cos \theta + i\sin \theta )}}$

$ \Rightarrow {e^{iz}} = {e^{ir\cos \theta + {i^2}r\sin \theta }}$

$ \Rightarrow {e^{iz}} = {e^{ir\cos \theta - r\sin \theta }}$

$ \Rightarrow {e^{iz}} = {e^{ - r\sin \theta }}\,.\,{e^{ir\cos \theta }}$

$\therefore$ $\arg ({e^{iz}}) = r\cos \theta $