[Om Sharma]

Solution

$\operatorname{Re}\left(\mathrm{a} \mathrm{z}^{2}+\mathrm{bz}\right)=\mathrm{a} \Rightarrow \mathrm{az}^{2}+\mathrm{bz}+\mathrm{a} \overline{\mathrm{z}}^{2}+\mathrm{b} \overline{\mathrm{z}}=2 \mathrm{a} \Rightarrow \mathrm{a}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{b}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{a}$

\[ \operatorname{Re}\left(\mathrm{b} \mathrm{z}^{2}+\mathrm{az}\right)=\mathrm{b} \Rightarrow \mathrm{~b} \mathrm{z}^{2}+\mathrm{az}+\mathrm{b} \overline{\mathrm{z}}^{2}+\mathrm{a} \overline{\mathrm{z}}=2 \mathrm{~b} \Rightarrow \mathrm{~b}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{a}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{~b} \]

\[ \begin{aligned} & (1) \times b-(2) \times(a) \Rightarrow\left(b^{2}-a^{2}\right)(z+\bar{z})=0 \Rightarrow(z+\bar{z})=0\left(a^{2} \neq b^{2}\right) \\ & (1) \times a-(2) \times(b) \Rightarrow\left(a^{2}-b^{2}\right)(z+\bar{z})=2\left(a^{2}-b^{2}\right)\left(a^{2} \neq b^{2}\right) \\ & z^{2}+\bar{z}^{2}=2 \Rightarrow(z+\bar{z})^{2}-2 z \bar{z}=2 \Rightarrow z \bar{z}=-1 \Rightarrow 1+1^{2}=-1 \Rightarrow \text { No solution But when } a=-b, \operatorname{Re}\left(a z^{2}-a z\right)=a \\ & \Rightarrow \operatorname{Re}\left(a\left(x^{2}-y^{2}+i 2 x y\right)-a(x+i y)\right)=a \Rightarrow a\left(x^{2}-y^{2}\right)-a x=a \Rightarrow x^{2}-y^{2}-x=1 \Rightarrow x^{2}-x-1=y^{2} \end{aligned} \]

For any real values of y there two values of x, hence infinite complex numbers are possible.

[Om Sharma]

Answer

Bonus

[Om Sharma]

Solution

$\frac{2 \mathrm{z}-3 \mathrm{i}}{2 \mathrm{z}+\mathrm{i}}$ is purely imaginary : $\frac{2 \mathrm{z}-3 \mathrm{i}}{2 \mathrm{z}+\mathrm{i}}+\frac{2 \overline{\mathrm{z}}-3 \mathrm{i}}{2 \overline{\mathrm{z}}-\mathrm{i}}=0$
$\mathrm{z}=\mathrm{x}+\mathrm{iy} \Rightarrow 4 \mathrm{x}^{2}+4 \mathrm{y}^{2}-4 \mathrm{y}-3=0$ Given that $\mathrm{x}+\mathrm{y}^{2}=0 \Rightarrow \mathrm{y}^{4}+\mathrm{y}^{2}-\mathrm{y}=3 / 4$

[Om Sharma]

Answer

D

[Om Sharma]

Solution

Let $\mathrm{a}=\mathrm{x}_{1}+\mathrm{iy}_{1} \mathrm{z}=\mathrm{x}+\mathrm{iy} \operatorname{Now} \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})>\operatorname{IM}(\overline{\mathrm{a}}+\mathrm{z}) \therefore \mathrm{x}_{1}+\mathrm{x}>-\mathrm{y}_{1}+\mathrm{y}$
$\mathrm{x}_{1}=2, \mathrm{y}_{1}=10, \mathrm{x}=-12, \mathrm{y}=0$ Given inequality is not valid for these values S1 is false. Now $\operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})<\operatorname{IM}(\overline{\mathrm{a}}+\mathrm{z})$ $\mathrm{x}_{1}+\mathrm{x}<-\mathrm{y}_{1}+\mathrm{y} \Rightarrow \mathrm{x}_{1}=-2, \mathrm{y}_{1}=-10, \mathrm{x}=12, \mathrm{y}=0$ Given inequality is not valid for these values. S2 is false

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$\alpha, \beta=\frac{-\sqrt{6} \pm \sqrt{6} \mathrm{i}}{2}=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$ Required expression $=\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)}=(\sqrt{3})^{8}=81$

[Om Sharma]

Answer

C

[Om Sharma]

Solution

$\mathrm{z}_{1}=\mathrm{x}_{1}+\mathrm{iy}_{1}, \mathrm{z}_{2}=\mathrm{x}_{2}+\mathrm{iy}_{2} \Rightarrow \operatorname{Re}\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)=\mathrm{x}_{1}+\mathrm{x}_{2}=0$
$\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ are of opposite sign, $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$ are of opposite sign

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$(1-\sqrt{3} \mathrm{i})^{200}=2^{199}(\mathrm{p}+\mathrm{iq}) \Rightarrow 2^{200}\left(\cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3}\right)^{200}=2^{199}(\mathrm{p}+\mathrm{iq})$

\[ \begin{aligned} & 2\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=p+i q \Rightarrow p=-1, q=-\sqrt{3} \\ & \alpha=p+q+q^{2}=2-\sqrt{3}, \beta=p-q+q^{2}=2+\sqrt{3} \\ & \alpha+\beta=4 \Rightarrow \alpha \cdot \beta=1 \text { equation } x^{2}-4 x+1=0 \end{aligned} \]

[Om Sharma]

Answer

A

[Om Sharma]

Solution

Let $\sin \frac{2 \pi}{9}+\mathrm{i} \cos \frac{2 \pi}{9}=\mathrm{z} \Rightarrow\left(\frac{1+\mathrm{z}}{1+\bar{z}}\right)^{3}=\left(\frac{1+\mathrm{z}}{1+\frac{1}{\mathrm{x}}}\right)^{3}=\mathrm{z}^{3}$

\[ \left(\mathrm{i}\left(\cos \frac{2 \pi}{9}-\mathrm{i} \sin \frac{2 \pi}{9}\right)\right)^{3}=-\mathrm{i}\left(\cos \frac{2 \pi}{3}-\mathrm{i} \sin \frac{2 \pi}{3}\right)=-\mathrm{i}\left(-\frac{1}{2}-\mathrm{i} \frac{\sqrt{3}}{2}\right) \Rightarrow-\frac{1}{2}(\sqrt{3}-\mathrm{i}) \]

[Om Sharma]

Answer

B

[Om Sharma]

Solution

$\mathrm{z}=1+\mathrm{i} \Rightarrow \mathrm{z}_{1}=\frac{1+\mathrm{i} \overline{\mathrm{z}}}{\overline{\mathrm{z}}(1-\mathrm{z})+1 / \mathrm{z}} \Rightarrow \mathrm{z}_{1}=\frac{1+\mathrm{i}(1-\mathrm{i})}{(1-\mathrm{i})(1-1-\mathrm{i})+\frac{1}{1+\mathrm{i}}}$

\[ \begin{aligned} & =\frac{1+\mathrm{i}-\mathrm{i}^{2}}{(1-\mathrm{i})(-\mathrm{i})+\frac{1-\mathrm{i}}{2}}=\frac{2+\mathrm{i}}{-\frac{3 \mathrm{i}-1}{2}}=\frac{4+2 \mathrm{i}}{-3 \mathrm{i}-1}=-\frac{(4+2 \mathrm{i})(3 \mathrm{i}-1)}{(3 \mathrm{i})^{2}-(1)^{2}}=-1+\mathrm{i} \\ & \operatorname{Arg}\left(\mathrm{z}_{1}\right)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \therefore \frac{12}{\pi} \arg \left(\mathrm{z}_{1}\right)=\frac{12}{\pi} \times \frac{3 \pi}{4}=9 \end{aligned} \]

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