[UPLIFTJEE]

Solution

Let $\mathrm{Z}=\mathrm{x}+\mathrm{iy}, \mathrm{x} \in \mathrm{R}, \mathrm{y} \in \mathrm{R} \Rightarrow \mathrm{x}-\mathrm{iy}=\mathrm{i}\left(\mathrm{x}^{2}-\mathrm{y}^{2}+(2 \mathrm{xy}) \mathrm{i}+\mathrm{x}\right)$
$\mathrm{x}=-2 \mathrm{xy} \ldots(1) \Rightarrow-\mathrm{y}=-\mathrm{y}^{2}+\mathrm{x}^{2}+\mathrm{x} \quad \ldots(2) \Rightarrow \mathrm{x}=0, \mathrm{y}=-\frac{1}{2}$ (from (1))
If $\mathrm{x} \neq 0$, then $\mathrm{y}=0,1 \Rightarrow$ If $\mathrm{y}=-\frac{1}{2}$, then $\mathrm{x}=\frac{1}{2},-\frac{3}{2} \Rightarrow \mathrm{Z}=0+\mathrm{i} 0,0+\mathrm{i}, \frac{1}{2}-\frac{\mathrm{i}}{2},-\frac{3}{2}-\frac{\mathrm{i}}{2}$

[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

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 $=\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)}=(\sqrt{3})^{8}=81$

[UPLIFTJEE]

Answer

C

[UPLIFTJEE]

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

[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

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

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

[UPLIFTJEE]

Answer

A

[UPLIFTJEE]

Solution

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

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

[UPLIFTJEE]

Answer

B

[UPLIFTJEE]

Solution

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

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

[UPLIFTJEE]

Answer

9

[UPLIFTJEE]

Solution

$4 \mathrm{x}^{4}+8 \mathrm{x}^{3}-17 \mathrm{x}^{2}-12 \mathrm{x}+9=4\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{x}-\mathrm{x}_{2}\right)\left(\mathrm{x}-\mathrm{x}_{3}\right)\left(\mathrm{x}-\mathrm{x}_{4}\right)$ Put $\mathrm{x}=2 \mathrm{i}$ and -2 i

\[ \begin{aligned} & 64-64 i+68-24 i+9=4\left(2 i-x_{1}\right)\left(2 i-x_{2}\right)\left(2 i-x_{3}\right)\left(2 i-x_{4}\right)=141-88 i \quad \ldots(1) \\ & 64+64 i+68+24 i+9=4\left(-2 i-x_{1}\right)\left(-2 i-x_{2}\right)\left(-2 i-x_{3}\right)\left(-2 i-x_{4}\right)=141+88 i \ldots(2) \text { product of }(1) \text { and }(2) \\ & 16\left(4+x_{1}{ }^{2}\right)\left(4+x_{2}{ }^{2}\right)\left(4+x_{3}{ }^{2}\right)\left(4+x_{4}{ }^{2}\right)=(141-88 i)(141+88 i) \Rightarrow\left(4+x_{1}{ }^{2}\right)\left(4+x_{2}{ }^{2}\right)\left(4+x_{3}{ }^{2}\right)\left(4+x_{4}{ }^{2}\right)=\frac{141^{2}+88^{2}}{16}=\frac{125}{16} \mathrm{~m} \Rightarrow \mathrm{~m}=221 \end{aligned} \]

[UPLIFTJEE]

Answer

221

[UPLIFTJEE]

Solution

$\alpha \& \beta$ are roots of $x^{2}-x+2=0 \Rightarrow \alpha+\beta=1, \alpha^{2}-\alpha+2=0, \alpha^{2}=\alpha-2$

\[ \begin{aligned} & \beta^{2}-\beta+2=0 \Rightarrow \beta^{2}=\beta-2 \Rightarrow \alpha^{6}+\alpha^{4}+\beta^{4}-5 \alpha^{2}=\alpha^{4}(\alpha-2)=\alpha^{4}(\alpha-2)+\alpha^{4}-5 \alpha^{2}+(\beta-2)^{2} \\ & =\alpha^{5}-\alpha^{4}-5 \alpha^{2}+\beta^{2}-4 \beta+4=\alpha^{3}(\alpha-2)-\alpha^{4}-5 \alpha^{2}+\beta-2-4 \beta+4=-2 \alpha^{3}-5 \alpha^{2}-3 \beta+2=-2 \alpha(\alpha-2)-5 \alpha^{2}-3 \beta+2 \\ & =-7 \alpha^{2}+4 \alpha-3 \beta+2=-7(\alpha-2)+4 \alpha-3 \beta+2=-3 \alpha-3 \beta+16=-3(\alpha+\beta)+16=13=-3(1)+16=13 \end{aligned} \]

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