Test Index

Complex Numbers Part 1

Section: Mathematics

© examsnet.com

Question : 21 of 100

Marks: +1, -0

Let

n

denote the number of solutions of the equation

z^{2}+3\overline{z}=0

z

is a complex number. Then the value of

\sum\limits_{k=0}^{\infty}\frac{1}{n^{k}}

[22 Jul 2021 Shift 2]

1
$\frac{4}{3}$
$\frac{3}{2}$
2

Solution:

z^{2}+3\overline{z}=0

Put

z=x+iy

\Rightarrow x^{2}-y^{2}+2ixy+3(x-iy)=0

\Rightarrow (x^{2}-y^{2}+3x)+i(2xy-3y)

=0+i0

\therefore x^{2}-y^{2}+3x=0

........(1)

2xy-3y=0

........(2)

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

Put

x=\frac{3}{2}

in equation (1)

\frac{9}{4}-y^{2}+\frac{9}{2}=0

y^{2}=\frac{27}{4}\Rightarrow y=\pm\frac{3\sqrt{3}}{2}

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

Put

y=0\Rightarrow x^{2}-0+3x=0

x=0,-3

\therefore (x,y)=(0,0),(-3,0)

\therefore

No of solutions

=n=4

\sum\limits_{K=0}^{\infty}\left(\frac{1}{n^{k}}\right)=\sum\limits_{K=0}^{\infty}\left(\frac{1}{4^{k}}\right)

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

=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}

© examsnet.com

Go to Question:

Prev Question

Next Question