x+5 =4i $\Rightarrow$ $x^2$+10x+25=-16

Now, $x^4$+9$x^3$+35$x^2$-x+4

=($x^2$+10x+41)($x^2$-x+4)-160=-160

$\frac{1-i}{1+i}$=$\frac{(1-i)(1-i)}{(1+i)(1-i)}$=$\frac{1+(i{)}^2-2i}{1+1}$=-i

Which can be written as cos$\frac{\pi }{2}$-isin$\frac{\pi }{2}$

According to condition, 3-i$x^2$ y = $x^2$  + y + 4i

⇒ $x^2$ + y = 3 And $x^2$y = -4 ⇒ x = ±2, y = -1

⇒  (x,y) = (2,-1) or (-2,-1)

Let $z=x+iy$
Here,  $z+\overline{z}=(x+iy)+(x-iy)=2x$ (Real)

And $z\overline{z}=(x+iy)(x-iy)=x^2+y^2$ (Real).

Let z = x+iy, then its coonjucate $\overline{z}$ = x-iy

Given that $z^2=(\overline{z}{)}^2$

⇒  $x^2$ - $y^2$ + 2ixy = $x^2$ - $y^2$ - 2ixy ⇒ 4ixy = 0

if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0  

Let z = x+iy, so that $\overline{z}$ = x - iy, therefore

$z^2+\overline{z}=0\iff (x^2-y^2+x)+i(2xy-y)$ = 0 

Equating real and imaginary parts , we get 

$x^2-y^2+x$ = 0 .......(i)

And 2xy - y = 0 ⇒ y = 0 or x = $\frac{1}{2}$

        if y = 0 , then (i) gives $x^2$ + x = 0 ⇒ x = 0 or 

                                                            x = -1

If x = $\frac{1}{2}$,

Then $x^2-y^2+x=0\Rightarrow y^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\Rightarrow y=\pm \frac{\sqrt{3}}{2}$

Hence, there are four solutions in all. 

Let  z = x+iy, $\overline{z}$  = x-iy

∴ z$\overline{z}$ = 0 ⇒ (x+iy)(x-iy) = 0 ⇒ $x^2+y^2$ = 0 

It is possible onle when x and y oth simultaneously zero 

i.e, z = 0 +0i = 0  

Here -1+$\sqrt{-3}$=r$e^{i\theta }$ $\Rightarrow$ -1+$i\sqrt{3}$=r$e^{i\theta }$

=rcos$\theta$=-1 And rsin$\theta$=$\sqrt{3}$

Hence tan$\theta$=-$\sqrt{3}$ $\Rightarrow$ tan$\theta$=tan$\frac{2\pi }{3}$.Hence $\theta$=$\frac{2\pi }{3}$

1=$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$=$|\frac{z_1\ast {\overline{z}}_1}{z_1}+\frac{z_2\ast {\overline{z}}_2}{z_2}+\frac{z_3\ast {\overline{z}}_3}{z_3}|$

(hence, ${|z_1|}^2$=1=$z_1$$\overline{z_1}$, etc)

$\Rightarrow$ |$z_1$+$z_2$+$z_3$|=|$\overline{z_1+z_2+z_3}$|= |$z_1$+$z_2$+$z_3$|

(hence,|$\overline{z_1}$|=|$z_1$|)

We have $\left|\frac{z-12}{z-8i}\right|$=$\frac{5}{3}$,  $\left|\frac{z-4}{z-8}\right|$=1

Let z=x+iy, then

$\left|\frac{z-12}{z-8i}\right|$=$\frac{5}{3}$ $\Rightarrow$ 3|z-12| = 5|z-8i|

$\Rightarrow$ $9(x-12{)}^2$+$9y^2$ = 25$x^2$+25$(y-8{)}^2$     $\cdots$$\cdots$(i)

$\left|\frac{z-4}{z-8}\right|$=1 $\Rightarrow$ |z-4| = |z-8|

$\Rightarrow$ $(x-4{)}^2$+$y^2$ = $y^2$+$(x-8{)}^2$ $\Rightarrow$ x=6

Putting x=6 in (i), we get $y^2$-25y+136=0

 y=17, 8

Hence z=6+17i or z=6+8i

Trick: Check it with options.