[Om Sharma]

Solution

$|\mathrm{z}+1|=\alpha \mathrm{z}+\beta(1+\mathrm{i})$, put $\mathrm{z}=\frac{1}{2}-2 \mathrm{i} \Rightarrow\left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i} \Rightarrow \sqrt{\frac{9}{4}+4}=\frac{5}{2}=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) \mathrm{i}$
Comparing real and imaginary parts $\frac{\alpha}{2}+\beta=\frac{5}{2}$ and $\beta=2 \alpha \Rightarrow \frac{\alpha}{2}+2 \alpha=\frac{5}{2} \Rightarrow \alpha=1, \beta=2$. Now, $\alpha+\beta=3$

[Om Sharma]

Solution

$1+10 \operatorname{Re}\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}\right)=0(\therefore 2 \operatorname{Re}(\mathrm{z})=\mathrm{z}+\overline{\mathrm{z}})$

\[ \begin{aligned} & \Rightarrow 1+5\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}\right)=0 \\ & \Rightarrow \frac{\left(2 \cos ^{2} \theta-3 \sin ^{2} \theta\right)+\left(2 \cos ^{2} \theta\right)-\left(3 \sin ^{2} \theta\right)}{\cos ^{2} \theta+9 \sin ^{2} \theta}=-\frac{1}{5} \\ & \Rightarrow \frac{2 \cos ^{2} \theta-3 \sin ^{2} \theta}{\cos ^{2} \theta+9 \sin ^{2} \theta}=-\frac{1}{10} \Rightarrow 20 \cos ^{2} \theta-30 \sin ^{2} \theta=-\cos ^{2} \theta-9 \sin ^{2} \theta \\ & \Rightarrow 21 \cos ^{2} \theta-21 \sin ^{2} \theta=0 \Rightarrow \cos 2 \theta=0 \text { As } \theta \in[0,2 \pi] \\ & \Rightarrow 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \end{aligned} \]

Now $\sum_{\theta \in \mathrm{A}} \theta^{2}=\frac{\pi^{2}}{16}+\frac{9 \pi^{2}}{16}+\frac{25 \pi^{2}}{16}+\frac{49 \pi^{2}}{16}=\frac{84 \pi^{2}}{16}=\frac{21 \pi^{2}}{4}$

[Om Sharma]

Answer

A

[Om Sharma]

Solution

$\mathrm{z}_{1}=\mathrm{e}^{-\mathrm{i} \pi 4}, \mathrm{z}_{2}=1, \mathrm{z}_{3}=\mathrm{e}^{\mathrm{i} \pi / 4} \Rightarrow\left|\mathrm{z}_{1} \overline{\mathrm{z}}_{2}+\mathrm{z}_{2} \overline{\mathrm{z}}_{3}+\mathrm{z}_{3} \overline{\mathrm{z}}_{1}\right|^{2}=\left|\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}} \times 1+1 \times \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{\mathrm{i} \frac{\pi}{4}} \times \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right|^{2}$

\[ \begin{aligned} & =\left|\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{\mathrm{i} \frac{\pi}{2}}\right|^{2}=\left|2 \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{i}\right|^{2}=|\sqrt{2}-\sqrt{2} \mathrm{i}+\mathrm{i}|^{2}=(\sqrt{2})^{2}+(1-\sqrt{2})^{2}=2+1+2-2 \sqrt{2}=5-2 \sqrt{2} \\ & \alpha=5, \beta=-2 \Rightarrow \alpha^{2}+\beta^{2}=29 \end{aligned} \]

[Om Sharma]

Answer

C

[Om Sharma]

Solution

Let $\mathrm{z}=\mathrm{x}+\mathrm{iy} ; \mathrm{A}:|\mathrm{z}-2-\mathrm{i}|=3 \Rightarrow|(\mathrm{x}-2)+(\mathrm{y}-1) \mathrm{i}|=3 \Rightarrow(\mathrm{x}-2)^{2}+(\mathrm{y}-1)^{2}=9$

\[ \begin{aligned} & B=\operatorname{Re}(z-i z)=2 \Rightarrow \operatorname{Re}((x+y)+i(y-x))=2 \Rightarrow x+y=2 \Rightarrow x=2-y \ldots(2) ; \text { On solving (1) and (2) } \\ & (2-y-2)^{2}+(y-1)^{2}=9 \Rightarrow 2 y^{2}-2 y-8=0 \Rightarrow y^{2}-y-4=0 \\ & y=\frac{1 \pm \sqrt{17}}{2}, x=2-y=\frac{3 \pm \sqrt{17}}{2} \Rightarrow z=\frac{3-\sqrt{17}}{2}+\left(\frac{1+\sqrt{17}}{2}\right) i ; z=\frac{3+\sqrt{17}}{2}+\left(\frac{1-\sqrt{17}}{2}\right) i \\ & \sum_{z \in S}|z|^{2}=\left(\left(\frac{3-\sqrt{17}}{2}\right)^{2}+\left(\frac{1+\sqrt{17}}{2}\right)^{2}\right)+\left(\left(\frac{3+\sqrt{17}}{2}\right)^{2}+\left(\frac{1-\sqrt{17}}{2}\right)^{2}\right)=\frac{2[9+17]}{4}+\frac{2[1+17]}{4}=13+9=22 \end{aligned} \]

[Om Sharma]

Answer

22

[Om Sharma]

Solution

$\omega_{1}=(8 \sin \theta+7 \cos \theta)+\mathrm{i}(\sin \theta+4 \cos \theta) ; \omega_{2}=(\sin \theta+4 \cos \theta)+\mathrm{i}(8 \sin \theta+7 \cos \theta)$

\[ \begin{aligned} & \omega_{1} \omega_{2}=(8 \sin \theta+7 \cos \theta)(\sin \theta+4 \cos \theta)+\mathrm{i}^{2}(\sin \theta+4 \cos \theta)(8 \sin \theta+7 \cos \theta)+\mathrm{i}\left[(\sin \theta+4 \cos \theta)^{2}+(8 \sin \theta+7 \cos \theta)^{2}\right] \\ & \omega_{1} \omega_{2}=0+\mathrm{i}\left(65 \sin ^{2} \theta+120 \sin \theta \cos \theta+65 \cos ^{2} \theta\right) \Rightarrow \omega_{1} \omega_{2}=0+\mathrm{i}(65+60 \sin 2 \theta)=\alpha+\mathrm{i} \beta \therefore \alpha+\beta=0+65+60 \sin 2 \theta \end{aligned} \]

We know that $-1 \leq \sin 2 \theta \leq 1 \Rightarrow \mathrm{p}=\alpha+\left.\beta\right|_{\text {max }}=65+60=125 ; \mathrm{q}=\alpha+\left.\beta\right|_{\text {min }}=65-60=5 \Rightarrow \mathrm{p}+\mathrm{q}=125+5=130$

[Om Sharma]

Answer

A

[Om Sharma]

Solution

$\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is purely real then $\operatorname{Im}\left(\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}\right)=0 \Rightarrow \frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\overline{\left(\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}\right)} \Rightarrow \frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{\overline{\mathrm{z}}+\mathrm{i}}{\overline{\mathrm{z}}-\mathrm{i}} \Rightarrow(\mathrm{z}-\mathrm{i})(\overline{\mathrm{z}}-\mathrm{i})=(\mathrm{z}+\mathrm{i})(\overline{\mathrm{z}}+\mathrm{i})$

\[ \Rightarrow \mathrm{z} \overline{\mathrm{z}}-\mathrm{zi}-\mathrm{i} \overline{\mathrm{z}}+\mathrm{i}^{2}=\mathrm{z} \overline{\mathrm{z}}+\mathrm{zi}+\mathrm{i} \overline{\mathrm{z}}+\mathrm{i}^{2} \Rightarrow 2 \mathrm{i}(\mathrm{z}+\overline{\mathrm{z}})=0 \Rightarrow \mathrm{z}+\overline{\mathrm{z}}=0 \]

let $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \mathrm{z}+\overline{\mathrm{z}}=\mathrm{x}+\mathrm{iy}+\mathrm{x}-\mathrm{iy}=0 \Rightarrow 2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0, \mathrm{z}=\mathrm{yi} \because|\mathrm{z}|=1 \Rightarrow|\mathrm{yi}|=1 \Rightarrow|\mathrm{y}|=1 \Rightarrow|\mathrm{y}|= \pm 1$
we get $\mathrm{x}=0, \mathrm{y}=1, \mathrm{z}=\mathrm{i}$ (accepted), $\mathrm{x}=0, \mathrm{y}=-1, \mathrm{z}=-\mathrm{i}$ but $\mathrm{z} \neq-\mathrm{i}$ (rejected) It has only one element in Set ∴ $\mathrm{S}_{1}$ is wrong
$\mathrm{S}_{2}: \frac{\mathrm{z}-1}{\mathrm{z}+1}$ is purely imaginary $\Rightarrow \operatorname{Re}\left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=0 \Rightarrow \frac{\mathrm{z}-1}{\mathrm{z}+1}+\overline{\left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)}=0 \Rightarrow \frac{\mathrm{z}-1}{\mathrm{z}+1}+\frac{\overline{\mathrm{z}}-1}{\overline{\mathrm{z}}+1}=0 \Rightarrow(\mathrm{z}-1)(\overline{\mathrm{z}}+1)+(\overline{\mathrm{z}}-1)(\mathrm{z}+1)=0$

\[ \Rightarrow z \bar{z}+z-\bar{z}-1+\bar{z} z-z+\bar{z}-1=0 \Rightarrow 2(z \bar{z}-1)=0 \Rightarrow z \bar{z}=1 \Rightarrow|z|^{2}=1 \Rightarrow|z|=1 \text { as given } z \neq-1 \]

$\mathrm{z}=\mathrm{re}^{\mathrm{i} \theta}, \mathrm{r}=1 \Rightarrow \mathrm{z}=\mathrm{e}^{\mathrm{i} \theta} \forall \theta \in \mathrm{R}-\{(2 \mathrm{n}+1) \pi\}$ It has $\infty$ solutions ∴ S-2 is True

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$\mathrm{z}=\frac{-\sqrt{6} \mathrm{i} \pm \sqrt{-6+12}}{2}=\frac{-\sqrt{6} \mathrm{i} \pm \sqrt{6}}{2}=\frac{\sqrt{6}}{2}( \pm 1-\mathrm{i})$

\[ \begin{aligned} & \alpha=-\frac{\sqrt{6}}{2}(1+i)=-\frac{\sqrt{6}}{2} \times \sqrt{2} e^{i \pi / 4}=-\sqrt{3} e^{i \pi / 4} \\ & \beta=\frac{\sqrt{6}}{2}(1-i)=\frac{\sqrt{6}}{2} \times \sqrt{2} e^{-i \pi / 4}=\sqrt{3} e^{-i \pi / 4} \\ & \sum_{z \in S} z^{8}=\alpha^{8}+\beta^{8}=81 \cdot e^{i 2 \pi}+81 \cdot e^{i(-2 \pi)}=81(1+1)=162 \end{aligned} \]

[Om Sharma]

Answer

A

[Om Sharma]

Solution

let $\mathrm{z}=\mathrm{x}+\mathrm{iy}, 4 \mathrm{z}^{2}+\overline{\mathrm{z}}=0 \Rightarrow 4(\mathrm{x}+\mathrm{iy})^{2}+(\mathrm{x}-\mathrm{iy})=0$

\[ \begin{aligned} & \Rightarrow 4\left(x^{2}-y^{2}+2 x y i\right)+x-i y=0 \Rightarrow 4 x^{2}-4 y^{2}+x+i y(8 x-1)=0+i 0 \Rightarrow 4 x^{2}-4 y^{2}+x=0 \text { and } y(8 x-1)=0 \\ & \text { Case I: } y=0 ; 4 x^{2}-0+x=0 \Rightarrow x=0,-\frac{1}{4} \Rightarrow \Rightarrow Z_{1}=0+i 0, Z_{2}=-\frac{1}{4}+i 0 \Rightarrow\left|z_{1}\right|^{2}=0 ;\left|z_{2}\right|^{2}=\frac{1}{16} \\ & \text { Case II: } 8 x-1=0 \Rightarrow x=\frac{1}{8} \Rightarrow 4 x^{2}-4 y^{2}+x=0 \Rightarrow \frac{4}{64}-y^{2}+\frac{1}{8}=0 \Rightarrow 4 y^{2}=\frac{3}{16} \Rightarrow y= \pm \frac{\sqrt{3}}{8} \\ & z_{3}=\frac{1}{8}+\frac{\sqrt{3}}{8} i, z_{4}=\frac{1}{8}-\frac{\sqrt{3}}{8} i \Rightarrow\left|z_{3}\right|^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} ;\left|z_{4}\right|^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \therefore \sum_{i=1}^{n}\left|z_{i}\right|^{2}=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16} \end{aligned} \]

[Om Sharma]

Answer

A

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