Let n denote the number of solutions of the equation z^2+3 z=0, where z is a complex number. Then the value of Σ _k=0^∞ 1 n^k is equal to

Given ${z}^{2}+3\bar{z}=0$z2+3zˉ=0

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

$${(x+iy)}^{2}+3(x-iy)=0$$(x+iy)2+3(x−iy)=0

$$\Rightarrow {x}^{2}+{i}^{2}{y}^{2}+2ixy+3x-3iy=0$$⇒x2+i2y2+2ixy+3x−3iy=0

We also, know that ${i}^{2}=-1,$i2=−1, hence, we get

$$\Rightarrow {x}^{2}-{y}^{2}+2ixy+3x-3iy=0$$⇒x2−y2+2ixy+3x−3iy=0

$$\Rightarrow ({x}^{2}-{y}^{2}+3x)+i(2xy-3y)=0+i0$$⇒(x2−y2+3x)+i(2xy−3y)=0+i0

On comparing the real and imaginary parts, we get

$${x}^{2}-{y}^{2}+3x=0\ldots (1)$$x2−y2+3x=0…(1)

And $2xy-3y=0\ldots (2)$2xy−3y=0…(2)

$$\Rightarrow y(2x-3)=0$$⇒y(2x−3)=0

$$\Rightarrow x=\frac{3}{2},y=0$$⇒x=23​,y=0

Put $x=\frac{3}{2}$x=23​ in equation $(1),$(1), we get $\frac{9}{4}-{y}^{2}+\frac{9}{2}=0$49​−y2+29​=0

$$\Rightarrow {y}^{2}=\frac{27}{4}$$⇒y2=427​

$$\Rightarrow y=\pm \frac{3\sqrt{3}}{2}.$$⇒y=±233​​.

$$\Rightarrow (x,y)=(\frac{3}{2},\frac{3\sqrt{3}}{2}),(\frac{3}{2},\frac{-3\sqrt{3}}{2}).$$⇒(x,y)=(23​,233​​),(23​,2−33​​).

Now, put $y=0,$y=0, in the equation $(1),$(1), we get ${x}^{2}+3x=0$x2+3x=0

$$\Rightarrow x=0,-3.$$⇒x=0,−3.

$$\therefore (x,y)=(0,0),(-3,0).$$∴(x,y)=(0,0),(−3,0).

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

Now, $\displaystyle \sum _{k=0}^{\infty }(\frac{1}{{n}^{k}})=\sum _{k=0}^{\infty }(\frac{1}{{4}^{k}})$k=0∑∞​(nk1​)=k=0∑∞​(4k1​)

$$=\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots$$=11​+41​+161​+641​+……

The above progression is a geometric progression, with first term $a=1$a=1 and common ratio $r=\frac{1}{4}$r=41​ and the sum of infinite terms of the geometric progression is $\frac{a}{1-r}$1−ra​

Thus, $\displaystyle \sum _{k=0}^{\infty }(\frac{1}{{n}^{k}})=\frac{1}{(1-\frac{1}{4})}$k=0∑∞​(nk1​)=(1−41​)1​

$$=\frac{1}{(\frac{3}{4})}=\frac{4}{3}.$$=(43​)1​=34​.