[Om Sharma]

Solution

$\mathrm{z}=\frac{\sqrt{3}+\mathrm{i}}{2}=\mathrm{e}^{\mathrm{i} \pi / 6} \Rightarrow \mathrm{z}^{201}=\left(\mathrm{e}^{\mathrm{i} \pi / 6}\right)^{201}=\mathrm{e}^{\mathrm{i} \frac{201 \pi}{6}}=\mathrm{e}^{\mathrm{i} \frac{67}{2} \pi}$

\[ \Rightarrow \mathrm{z}^{201}=\mathrm{e}^{\mathrm{i}\left(32 \pi+\frac{3}{2} \pi\right)}=\mathrm{e}^{\mathrm{i} \frac{3 \pi}{2}}=-\mathrm{i} \text { Now }\left(\mathrm{z}^{201}-\mathrm{i}\right)^{8}=(-\mathrm{i}-\mathrm{i})^{8}=(-2 \mathrm{i})^{8}=(-1)^{8} 2^{8} \mathrm{i}^{8}=2^{8}=256 \]

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$|\mathrm{z}|=|1+\mathrm{i}\|1+2 \mathrm{i}\| 1+3 \mathrm{i}| \cdots \cdots|1+\mathrm{ni}| \Rightarrow|\mathrm{z}|^{2}=(1+1)(1+4)(1+9) \ldots\left(1+\mathrm{n}^{2}\right)$

\[ |z|^{2}=44200=2 \times 5 \times 10 \times 17 \times 26 \Rightarrow|z|^{2}=(1+1)\left(1+2^{2}\right)\left(1+3^{2}\right)\left(1+4^{2}\right)\left(1+5^{2}\right) \]

From (1) $\therefore \mathrm{n}=5$

[Om Sharma]

Answer

5

[Om Sharma]

Solution

$\frac{z^{2}+3 i}{z-2+i}=2+3 i \Rightarrow z^{2}+3 i=z(2+3 i)-7-4 i \Rightarrow z^{2}-z(2+3 i)+7+7 i=0$ has roots $z_{1}$ and $z_{2}$

\[ \mathrm{z}_{1}+\mathrm{z}_{2}=2+3 \mathrm{i}, \mathrm{z}_{1} \mathrm{z}_{2}=7+7 \mathrm{i} \]

Required sum is $\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)^{2}-2 \mathrm{z}_{1} \mathrm{z}_{2}=(2+3 \mathrm{i})^{2}-2(7+7 \mathrm{i})=4-9+12 \mathrm{i}-14-14 \mathrm{i}=-19-2 \mathrm{i}$

[Om Sharma]

Answer

C

[Om Sharma]

Solution

$50\left(\frac{2 \mathrm{x}(1-3 \mathrm{i})}{10}-\frac{\mathrm{y}(1+2 \mathrm{i})}{5}\right)=31+17 \mathrm{i}\Rightarrow 10 \mathrm{x}-(30 \mathrm{x}) \mathrm{i}-10 \mathrm{y}-20 \mathrm{yi}=31+17 \mathrm{i}$

\[ \begin{aligned} & 10(x-y)=31 \ldots(1) \text { and }-30 x-20 y=17 \\ & \Rightarrow 30 x-30 y-30 x-20 y=93+17 \Rightarrow y=-\frac{110}{50}=-2.2, x=0.9 \therefore 10(x-3 y)=75 \end{aligned} \]

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$|\mathrm{z}+2|=|\mathrm{z}-2|$ let $\mathrm{z}=\mathrm{x}+\mathrm{iy}$

\[ \begin{aligned} & |x+2+i y|=|x-2+i y| \Rightarrow(x+2)^{2}+y^{2}=(x-2)^{2}+y^{2} \Rightarrow x=0 \\ & \Rightarrow z=0+i y ; y \in R \\ & \arg \left(\frac{3+i y}{i y-i}\right)=\arg \left(\frac{y}{y-1}+\frac{3 i}{1-y}\right)=\frac{\pi}{4} \\ & \Rightarrow \frac{y}{y-1}=\frac{3}{1-y}>0 \Rightarrow y=-3 \Rightarrow z=-3 i \Rightarrow|z|^{2}=9 \end{aligned} \]

[Om Sharma]

Answer

A

[Om Sharma]

Solution

$\mathrm{z}^{2}+4 \mathrm{z}+16=0 \Rightarrow(\mathrm{z}+2)^{2}=-12 \Rightarrow \mathrm{z}=-2 \pm 2 \sqrt{3} \mathrm{i}$

\[ \begin{aligned} & \Rightarrow S=\{-2+2 \sqrt{3} i,-2-2 \sqrt{3} i\} \\ & \therefore \sum_{z \in S}|z+\sqrt{3} i|^{2}=|-2+2 \sqrt{3} i+\sqrt{3} i|^{2}+|-2-2 \sqrt{3} i+\sqrt{3} i|^{2} \\ & =|-2+3 \sqrt{3} i|^{2}+|-2-\sqrt{3} i|^{2}=4+27+4+3=38 \end{aligned} \]

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$\mathrm{z}^{2}+4 \mathrm{z}-(1+12 \mathrm{i})=0$.

\[ \begin{aligned} & \Rightarrow \mathrm{z}=\frac{-4 \pm \sqrt{16+4(1+12 \mathrm{i})}}{2}=-2 \pm \sqrt{5+12 \mathrm{i}} \\ & \because \sqrt{5+12 \mathrm{i}}=\mathrm{x}+\mathrm{iy} \text { where } \mathrm{x}, \mathrm{y} \in \mathrm{R} \\ & \Rightarrow 5+12 \mathrm{i}=\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xyi} \\ & \Rightarrow 2 \mathrm{xy}=12 \Rightarrow \mathrm{y}=\frac{6}{\mathrm{x}} \& \mathrm{x}^{2}-\mathrm{y}^{2}=5 \\ & \Rightarrow \mathrm{x}^{2}-\frac{36}{\mathrm{x}^{2}}=5 \Rightarrow \mathrm{x}^{4}-5 \mathrm{x}^{2}-36=0 \\ & \Rightarrow\left(\mathrm{x}^{2}-9\right)\left(\mathrm{x}^{2}+4\right)=0 \Rightarrow \mathrm{x}^{2}=-4 \Rightarrow \mathrm{x}= \pm 2 \mathrm{i} \text { (rejected) } \\ & \Rightarrow \mathrm{x}^{2}=9 \Rightarrow \mathrm{x}= \pm 3 ; \mathrm{y}=\frac{6}{\mathrm{x}}= \pm 2 \\ & \because \sqrt{5+12 \mathrm{i}}= \pm(3+2 \mathrm{i}) \end{aligned} \]

\[ \begin{aligned} & \Rightarrow z=-2 \pm(3+2 i) \Rightarrow z=1+2 i,-5-2 i \\ & \therefore\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=(1+4)+(25+4)=5+29=34 \end{aligned} \]

[Om Sharma]

Answer

D

[Om Sharma]

Solution

Put $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy}$

\[ \begin{aligned} & \mathrm{z}(\overline{\mathrm{z}}+2+\mathrm{i})+\mathrm{k}(2+3 \mathrm{i})=0 \\ & \Rightarrow(\mathrm{x}+\mathrm{iy})(\mathrm{x}-\mathrm{iy}+2+\mathrm{i})+2 \mathrm{k}+3 \mathrm{ki}=0 \\ & \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-\mathrm{y}+2 \mathrm{k}+(\mathrm{x}+2 \mathrm{y}+3 \mathrm{k}) \mathrm{i}=0+\mathrm{i} 0 \\ & \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-\mathrm{y}+2 \mathrm{k}=0 \text { and } \mathrm{x}+2 \mathrm{y}+3 \mathrm{k}=0 \Rightarrow \mathrm{x}=-2 \mathrm{y}-3 \mathrm{k} \end{aligned} \]

eliminating x from first equation

\[ \begin{aligned} & 5 \mathrm{y}^{2}+(12 \mathrm{k}-5) \mathrm{y}+9 \mathrm{k}^{2}-4 \mathrm{k}=0 \text { since } \mathrm{y} \in \mathrm{R} \text { so } \mathrm{D} \geq 0 \\ & (12 \mathrm{k}-5)^{2}-4.5\left(9 \mathrm{k}^{2}-4 \mathrm{k}\right) \geq 0 \Rightarrow 36 \mathrm{k}^{2}+40 \mathrm{k}-25 \leq 0 \text { then } \mathrm{k} \in[\alpha, \beta] \\ & \Rightarrow \alpha+\beta=-\frac{10}{9} \Rightarrow 9(\alpha+\beta)=-10 \end{aligned} \]

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