Complex Numbers 2 Question 41
42. Find all non-zero complex numbers $z$ satisfying $\bar{z}=i z^{2}$.
(1996, 2M)
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Solution:
- Let
$$ \begin{aligned} z & =x+i y . \\ \bar{z} & =i z^{2} \\ (\overline{x+i y}) & =i(x+i y)^{2} \\ x-i y & =i\left(x^{2}-y^{2}+2 i x y\right) \\ x-i y & =-2 x y+i\left(x^{2}-y^{2}\right) \end{aligned} $$
$$ \begin{aligned} & \text { Given, } \\ & \Rightarrow \\ & \Rightarrow \end{aligned} $$
NOTE It is a compound equation, therefore we can generate from it more than one primary equations.
On equating real and imaginary parts, we get
$$ \begin{array}{rlrl} & & x & =-2 x y \text { and }-y=x^{2}-y^{2} \\ \Rightarrow & & x+2 x y & =0 \text { and } x^{2}-y^{2}+y=0 \\ \Rightarrow & x(1+2 y) & =0 \\ \Rightarrow & x & =0 \text { or } y=-1 / 2 \end{array} $$
When $x=0, x^{2}-y^{2}+y=0 \Rightarrow 0-y^{2}+y=0$
$\Rightarrow \quad y(1-y)=0 \quad \Rightarrow \quad y=0 \quad$ or $\quad y=1$
When, $y=-1 / 2, x^{2}-y^{2}+y=0$
$$ \begin{array}{rlrl} \Rightarrow & & x^{2}-\frac{1}{4}-\frac{1}{2} & =0 \Rightarrow x^{2}=\frac{3}{4} \\ \Rightarrow & x & = \pm \frac{\sqrt{3}}{2} \end{array} $$
Therefore, $z=0+i 0,0+i ; \pm \frac{\sqrt{3}}{2}-\frac{i}{2}$
$\Rightarrow \quad z=i, \pm \frac{\sqrt{3}}{2}-\frac{i}{2}$
