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, .........

     Let z = x + iy, $\overline{z}$ = x -iy  and  $z^{-1}=\frac{1}{x+iy}$
⇒ $\overline{z^{-1}}=\frac{x+iy}{x^2+y^2}$ ; ∴ $\overline{z^{-1}}\overline{z}=\frac{x+iy}{x^2+y^2}(x-iy)$  = 1 

sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosx and cos2x=sin2x

Or     tan x = 1   ⇒ x   = $\frac{\pi }{4}$,$\frac{5\pi }{4}$,$\frac{9\pi }{4}$, ..........            ...............(i)

And  tan 2x = 1 ⇒ 2x = $\frac{\pi }{4}$,$\frac{5\pi }{4}$,$\frac{9\pi }{4}$, ..........

or                              x  =  $\frac{\pi }{8}$,$\frac{5\pi }{8}$,$\frac{9\pi }{8}$, ..........           ...............(ii)

There exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.

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)

$r^2+\frac{1}{r^2}-2=a^2\Rightarrow r-\frac{1}{r}$=$a\Rightarrow r$=$\frac{a+\sqrt[2]{2a^2+4}}{2}$

 $\sqrt{3}+i$=(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

Since1-i = $\sqrt{2}$ $[cos\frac{\pi }{4}-isin\frac{\pi }{4}]$, |1-i|

$\therefore$ $|1-i{|}^x$=$2^x$ $\Rightarrow$ ${\left(\sqrt{2}\right)}^x$=$2^x$ $\Rightarrow$ $2^{x/2}$ =$2^x$

$\Rightarrow$ $\frac{x}{2}$=x then x=0.

Therefore, the number of non-zero integral solutions is nil or Zero.

Let $\sqrt{3-4i}$=x+iy $\Rightarrow$ 3-4i= $x^2$-$y^2$+2ixy

$\Rightarrow$ $x^2$-$y^2$=3, 2xy=-4   ..........(i)

$\Rightarrow$ ${(x^2+y^2)}^2$=${(x^2-y^2)}^2$+4$x^2$$y^2$=$(3{)}^2$+$(-4{)}^2$=25

$\Rightarrow$ ${(x^2+y^2)}^2$=5           ............(ii)

From equation (i) and (ii) $\Rightarrow$ $x^2$=4 $\Rightarrow$x=$\pm$2,

$y^2$=1 $\Rightarrow$y=$\pm$1.Hence the square root of (3-4i)

is $\pm$(2-i).

      $\frac{c+i}{c-i}$ = a+ib     .........(i)

∴   $\frac{c-i}{c+i}$ = a-ib     .........(ii)

Multiplying (i) and (ii), we get 

$\frac{c^2+1}{c^2+1}=a^2+b^2\Rightarrow a^2+b^2=1$.

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 = $\frac{13}{10}-i\frac{9}{10}$.

we have

(1+i)(1+2i)(1+3i)......(1+ni) = a+ib   .......(i)

$\Rightarrow$ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)