Mathematics - Complex Number Question with Solution | TestHub

Given $z^2+3\overline{z}=0$

Put $z=x+iy,$ then we know that $\overline{z}=x-iy,$ hence, we have

${\left(x+iy\right)}^2+3\left(x-iy\right)=0$

$\Rightarrow x^2+i^2{y}^2+2ixy+3x-3iy=0$

We also, know that $i^2=-1,$ hence, we get

$\Rightarrow x^2-y^2+2ixy+3x-3iy=0$

$\Rightarrow \left(x^2-y^2+3x\right)+i(2xy-3y)=0+i0$

On comparing the real and imaginary parts, we get

$x^2-y^2+3x=0 \dots \left(1\right)$

And $2xy-3y=0 \dots \left(2\right)$

$\Rightarrow y\left(2x-3\right)=0$

$\Rightarrow x=\frac{3}{2}, y=0$

Put $x=\frac{3}{2}$ in equation $\left(1\right),$ we get $\frac{9}{4}-y^2+\frac{9}{2}=0$

$\Rightarrow y^2=\frac{27}{4}$

$\Rightarrow y=\pm \frac{3\sqrt{3}}{2}.$

$\Rightarrow \left(x, y\right)=\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right), \left(\frac{3}{2}, \frac{-3\sqrt{3}}{2}\right).$

Now, put $y=0,$ in the equation $\left(1\right),$ we get $x^2+3x=0$

$\Rightarrow x=0, -3.$

$\therefore \left(x, y\right)=\left(0, 0\right), \left(-3, 0\right).$

$\therefore$ No of solutions$=n=4$

Now, $\sum _{k=0}^{\infty }\left(\frac{1}{n^k}\right)=\sum _{k=0}^{\infty }\left(\frac{1}{4^k}\right)$

$=\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots \dots$

The above progression is a geometric progression, with first term $a=1$ and common ratio $r=\frac{1}{4}$ and the sum of infinite terms of the geometric progression is $\frac{a}{1-r}$

Thus, $\sum _{k=0}^{\infty }\left(\frac{1}{n^k}\right)=\frac{1}{\left(1-{\displaystyle \frac{1}{4}}\right)}$

$=\frac{1}{\left({\displaystyle \frac{3}{4}}\right)}=\frac{4}{3}.$