[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

$Z=\frac{(1+i \cos \theta)(1+2 i \cos \theta)}{(1-2 i \cos \theta)(1+2 i \cos \theta)} \Rightarrow Z=\frac{1-2 \cos ^{2} \theta+3 i \cos \theta}{1+4 \cos ^{2} \theta}$ is purely imaginary
So, $\operatorname{Re}(\mathrm{Z})=0 \Rightarrow \frac{1-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}=0 \Rightarrow \cos ^{2} \theta=\frac{1}{2} \Rightarrow \theta=-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ sum of all solutions $=3 \pi$

[UPLIFTJEE]

Answer

D

[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 $x^{2}-x+2=0 \Rightarrow \alpha+\beta=1, \alpha^{2}-\alpha+2=0, \alpha^{2}=\alpha-2$

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

[UPLIFTJEE]

Answer

13

[UPLIFTJEE]

Answer

B

[UPLIFTJEE]

Solution

$\mathrm{z}_{1}+\mathrm{z}_{2}=5$ and $\mathrm{z}_{1}^{3}+\mathrm{z}_{2}^{3}=20+15 \mathrm{i}$

\[ \begin{aligned} & z_{1}^{3}+z_{2}^{3}=\left(z_{1}+z_{2}\right)^{3}-3 z_{1} z_{2}\left(z_{1}+z_{2}\right) \Rightarrow z_{1}^{3}+z_{2}^{3}=125-3 z_{1} \cdot z_{2}(5) \\ & \Rightarrow 20+15 i=125-15 z_{1} z_{2} \Rightarrow 3 z_{1} z_{2}=25-4-3 i=21-3 i \Rightarrow z_{1} \cdot z_{2}=7-i \\ & \Rightarrow\left(z_{1}+z_{2}\right)^{2}=25 \Rightarrow z_{1}^{2}+z_{2}^{2}+2 z_{1} z_{2}=25 \Rightarrow z_{1}^{2}+z_{2}^{2}=25-2(7-i)=11+2 i \\ & \left(z_{1}^{2}+z_{2}^{2}\right)^{2}=121-4+44 i \Rightarrow z_{1}^{4}+z_{2}^{2}+2\left(z_{1} z_{2}\right)^{2}=117+44 i \Rightarrow z_{1}^{4}+z_{2}^{4}+2(7-i)^{2}=117+44 i \\ & \Rightarrow z_{1}^{4}+z_{2}^{4}=117+44 i-2(49-1-14 i)=21+72 i=3(7+24 i) \Rightarrow\left|z_{1}^{4}+z_{2}^{4}\right|=3 \sqrt{49+576}=3 \sqrt{625}=3 \times 25=75 \end{aligned} \]

[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

$\mathrm{x}^{2}-\sqrt{6} \mathrm{x}+3=0 \Rightarrow \mathrm{x}=\frac{\sqrt{6}}{2}(1 \pm \mathrm{i})=\frac{\sqrt{3}(1 \pm \mathrm{i})}{\sqrt{2}}$. Roots are $\alpha=\sqrt{3}\left(\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right)$

\[ \because \frac{\alpha^{99}}{\beta}+\alpha^{98}=\frac{\alpha^{98}(\alpha+\beta)}{\beta}=\frac{\left(\sqrt{3} \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right)^{98}\left[\sqrt{3} \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}+\sqrt{3} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right]}{\sqrt{3} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}}=3^{49} \mathrm{e}^{\mathrm{i} \frac{49 \pi}{2}}\left(2 \cos \frac{\pi}{4}\right) \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}=3^{49} \frac{(\mathrm{i})}{\sqrt{2}} 2\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)=3^{49}(-1+\mathrm{i})=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib}) \]

we get $\mathrm{n}=49, \mathrm{a}=-1, \mathrm{~b}=1 \therefore \mathrm{n}+\mathrm{a}+\mathrm{b}=49-1+1=49$

[UPLIFTJEE]

Answer

49

[UPLIFTJEE]

Solution

$\mathrm{z}=2-\mathrm{i}\left(2 \tan \frac{5 \pi}{8}\right)=\mathrm{x}+\mathrm{iy}$ (let)

\[ \begin{aligned} & |z|=r=\sqrt{x^{2}+y^{2}}=\sqrt{(2)^{2}+\left(2 \tan \frac{5 \pi}{8}\right)^{2}}=\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right|=2 \sec \frac{3 \pi}{8}\left(\sec \frac{5 \pi}{8}<0\right) \\ & \arg (z)=\theta=\tan ^{-1}\left(\frac{y}{x}\right) \text { as } x, y>0=\tan ^{-1}\left(\frac{-2 \tan \frac{5 \pi}{8}}{2}\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right)=\tan ^{-1}\left(\tan \frac{3 \pi}{8}\right)=\frac{3 \pi}{8} \end{aligned} \]

[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

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

\[ \Rightarrow x^{2}-y^{2}-2 i x y+\sqrt{x^{2}+y^{2}}=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}-2 x y i=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}=0 \text { and } x y=0 \]

Case $\mathrm{I}: \mathrm{x}=0 \quad$ Case II : $\mathrm{y}=0$

\[ \begin{array}{ll} -\mathrm{y}^{2}+|\mathrm{y}|=0 \Rightarrow|\mathrm{y}|=|\mathrm{y}|^{2} & \mathrm{x}^{2}+|\mathrm{x}|=0 \\ \Rightarrow|\mathrm{y}|(|\mathrm{y}|-1)=0 & \Rightarrow \mathrm{x}=0 \\ \Rightarrow \mathrm{y}=0, \pm 1 & \Rightarrow \mathrm{z}=0 \end{array} \]

$\mathrm{z}=0, \mathrm{i},-\mathrm{i}$ are roots $\alpha=\mathrm{i}-\mathrm{i}=0 ; \beta=\mathrm{i}(-\mathrm{i})=1 \Rightarrow 4\left(\alpha^{2}+\beta^{2}\right)=4(0+1)=4$

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