Answer: Option A. -> 2
:
A
${(\frac{i-1}{i+1}\times \frac{i-1}{i-1})}^n={\left(\frac{-2i}{-2}\right)}^n=i^n$
Hence, to make the real number the least positive integar is 2.

Answer: Option C. -> 0
:
C
cos$\alpha$+cos$\beta$+cos$\gamma$=0 ......(i)
sin$\alpha$+sin$\beta$+sin$\gamma$=0 .......(ii)
Let a=cos$\alpha$+isin$\alpha$, b=cos$\beta$+isin$\beta$
c=cos$\gamma$+isin$\gamma$
$\Rightarrow$ a+b+c=0 .......(iii)
$\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$
=${[cos\alpha +isin\alpha ]}^{-1}$+${[cos\beta +isin\beta ]}^{-1}$+${[cos\gamma +isin\gamma ]}^{-1}$
=cos$\alpha$-isin$\alpha$+cos$\beta$-isin$\beta$+cos$\gamma$-isin$\gamma$
$\Rightarrow$ ab+bc+ca=0 .....(iv) [by (i) and (ii)]
Squaring both sides of equations (iii),
We get $a^2$+$b^2$+$c^2$+2ab+2bc+2ca=0
or $a^2$+$b^2$+$c^2$ [by (iv)]
${[cos\alpha +isin\alpha ]}^2$+${[cos\beta +isin\beta ]}^2$+${[cos\gamma +isin\gamma ]}^2$=0
$\Rightarrow$ (cos2$\alpha$+cos2$\beta$+cos2$\gamma$) + i(sin2$\alpha$+sin2$\beta$+sin2$\gamma$)
Equating the real part to zero, we have
$\Rightarrow cos2\alpha +cos2\beta +cos2\gamma =0$

Answer: Option C. -> 3+5i
:
C
z = 3+5i ,$\overline{z}$ = 3-5i
⇒$z^3=(3+5i{)}^3=3^3+(5i{)}^3+\mathrm{3.3.5}i(3+5i)$
= -198+10i
Hence, $z^3+\overline{z}$+ 198 = 10i-198+3-5i+198 = 3+5i .

Answer: Option B. -> nπ+π6, nϵI
:
B
$\sqrt{3}+i$=(a+ib)(c+id)
$\therefore ac-bd$=$\sqrt{3}$ and ad+bc=1
Now ${\tan }^{-1}\left(\frac{b}{a}\right)+{\tan }^{-1}\left(\frac{d}{c}\right)$
= ${\tan }^{-1}\left(\frac{\frac{a}{b}+\frac{d}{c}}{1-\frac{b}{a}.\frac{d}{c}}\right)$= ${\tan }^{-1}\left(\frac{bc+ad}{ac-bd}\right)$=${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
=n$\pi$+$\frac{\pi }{6}$, n$ϵ$I

Answer: Option D. -> 1
:
D
$\frac{b}{c}$=$\frac{cos\beta +isin\beta }{cos\gamma +isin\gamma }\times \frac{cos\gamma -isin\gamma }{cos\gamma -isin\gamma }$
$\Rightarrow \frac{b}{c}=cos(\beta -\gamma )+isin(\beta -\gamma ).....\left(i\right)$
Similarly,$\frac{c}{a}=cos(\gamma -\alpha )+isin(\gamma -\alpha ).......\left(ii\right)$
And $\frac{a}{b}=cos(\alpha -\beta )+isin(\alpha -\beta ).......\left(ii\right)$
Equating real and imaginary parts,
$cos(\beta -\gamma )+cos(\gamma -\alpha )+cos(\alpha -\beta )=1$

Answer: Option B. -> 32
:
B
cos$\alpha$+cos$\beta$+cos$\gamma$ = 0............(i)
sin$\alpha$+sin$\beta$+sin$\gamma$ =0...........(ii)
Let a=cos$\alpha$+isin$\alpha$, b=cos$\beta$+isin$\beta$
c=cos$\gamma$+isin$\gamma$
$\Rightarrow$ a+b+c=0[By (i) and (ii)]....(iii)
$\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$
=cos$\alpha$-isin$\alpha$+cos$\beta$-isin$\beta$+cos$\gamma$-isin$\gamma$
$\Rightarrow$ ab+bc+ca=0 ......(iv) [by (i) and (ii)]
Squaring both sides of equation (iii),
we get $a^2$+$b^2$+$c^2$+2ab+2bc+2ca=0
or $a^2$+$b^2$+$c^2$=0 [by (iv)]
$[cos\alpha +isin\alpha {]}^2$ + $[cos\beta +isin\beta {]}^2$ + $[cos\gamma +isin\gamma {]}^2$ = 0
$\Rightarrow$ (cos2$\alpha$+cos2$\beta$+cos2$\gamma$) + i(sin2$\alpha$+sin2$\beta$+sin2$\gamma$)=0
Separation of real and imaginary part,
$\Rightarrow$ cos2$\alpha$+cos2$\beta$+cos2$\gamma$=0 .........(v)
And sin2$\alpha$+sin2$\beta$+sin2$\gamma$ =0 .........(iv)
$\Rightarrow$ 1-2$si{n}^2$$\alpha$+1-2$si{n}^2$$\beta$+1-2$si{n}^2$$\gamma$=0 [By eq.(v)]
Therefore, $si{n}^2$$\alpha$+$si{n}^2$$\beta$+$si{n}^2$$\gamma$= $\frac{3}{2}$

Answer: Option C. -> 1 if n is odd, i if n is even
:
C
Let z =$i^{[1+3+5+......+(2n+1\left)\right]}$
Clearly series is A.P with common difference = 2
∵$T_n$ = 2n-1 And $T_{n+1}$ = 2n+1
So, number of terms in A.P. = n+1
Now, $s_{n+1}=\frac{n+1}{2}$ [2.1+(n+1-1)2]
$\Rightarrow$ $S_{n+1}=\frac{n+1}{2}[2+2n]=(n+1{)}^2$ i.e., $i^{(n+1{)}^2}$
Now put n = 1,2,3,4,5,........
n = 1, z =$i^4$ = 1, n = 2, z = $i^5$= -1 ,
n = 3, z =$i^8$ = 1, n = 4, z = $i^{10}$= -1 ,
n = 5, z =$i^{12}$ = 1, .........

Answer: Option C. -> 12(2√(a2+4)+a)
:
C
let z=r (cos$\theta$+isin$\theta$)
Then $|z+\frac{1}{z}|$=a $\Rightarrow$ ${|z+\frac{1}{z}|}^2$=$a^2$
$\Rightarrow$ $r^2$+$\frac{1}{r^2}$+2cos$\theta$ = $a^2$ $\cdots$ $\cdots$(i)
Differentiating w.r.t $\theta$ we get
2r$\frac{dr}{d\theta }$-$\frac{2}{r^3}$$\frac{dr}{d\theta }$-4sin2$\theta$
Putting $\frac{dr}{d\theta }$=0, we get $\theta$=0,$\frac{\pi }{2}$
r is maximum for $\theta$ = 0, $\frac{\pi }{2}$, therefore from (i)
$r^2+\frac{1}{r^2}-2=a^2\Rightarrow r-\frac{1}{r}$=$a\Rightarrow r$=$\frac{a+\sqrt[2]{2a^2+4}}{2}$

Answer: Option C. -> 1310+i(−910)
:
C
z = $\frac{(2+i{)}^2}{3+i}=\frac{3+4i}{3+i}\times \frac{3-i}{3-i}=\frac{13}{10}+i\frac{9}{10}$
Conjugate = $\frac{13}{10}-i\frac{9}{10}$.

Answer: Option C. -> 1 if n is odd, i if n is even
:
C
Let z =$i^{[1+3+5+......+(2n+1\left)\right]}$
Clearly series is A.P with common difference = 2
∵$T_n$ = 2n-1 And $T_{n+1}$ = 2n+1
So, number of terms in A.P. = n+1
Now, $s_{n+1}=\frac{n+1}{2}$ [2.1+(n+1-1)2]
$\Rightarrow$ $S_{n+1}=\frac{n+1}{2}[2+2n]=(n+1{)}^2$ i.e., $i^{(n+1{)}^2}$
Now put n = 1,2,3,4,5,........
n = 1, z =$i^4$ = 1, n = 2, z = $i^5$= -1 ,
n = 3, z =$i^8$ = 1, n = 4, z = $i^{10}$= -1 ,
n = 5, z =$i^{12}$ = 1, .........