3
$\begingroup$

Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.

My approach is as follow

$z = r{e^{i\theta }}$& $\overline z = r{e^{ - i\theta }}$

$r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = i\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right) \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = {e^{i\frac{\pi }{2}}}\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right)$

$ \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = \left( {r{e^{ - i\left( {\theta - \frac{\pi }{2}} \right)}} + {r^2}{e^{i\left( {2\theta + \frac{\pi }{2}} \right)}}} \right)$

$ \Rightarrow r\left( {\cos \theta - i\sin \theta } \right) - {r^2}\left( {\cos 2\theta + i\sin 2\theta } \right) = \left( {r\left( {\sin \theta + i\cos \theta } \right) + {r^2}\left( { - \sin 2\theta + i\cos 2\theta } \right)} \right)$

$ \Rightarrow r\cos \theta - {r^2}\cos 2\theta - r\sin \theta + {r^2}\sin 2\theta - i\left( {r\sin \theta - {r^2}\sin 2\theta - r\cos \theta - {r^2}\cos 2\theta } \right) = 0$

Not able to proceed further

$\endgroup$
1
  • $\begingroup$ @insipidintegrator I know this question is addressed to the asker, but does he/she understand it ? Personnaly, I don't catch its meaning. $\endgroup$
    – Jean Marie
    Commented Aug 31, 2022 at 8:45

3 Answers 3

6
$\begingroup$

First of all, clearly $z = 0$ works, so let's just assume $z \neq 0$.

Let $z = r e^{i \theta}$. We have:

\begin{align} \bar{z} - z^2 &= i(\bar{z} + z^2) \\ (1 - i)\bar{z} &= (1 + i)z^2 \\ \frac{1 - i}{1 + i} &= \frac{z^2}{\bar{z}} \\ e^{-i \pi/2} &= r e^{3i\theta} \end{align}

This shows that $r = 1$ and $\theta = -\frac{\pi}{6} + \frac{2 \pi}{3} n$ for some $n \in \mathbb{Z}$.

In summary, the solutions are $0, e^{-i\pi/6}, e^{i 3\pi / 6}$ and $e^{i 7\pi / 6}$.

$\endgroup$
1
$\begingroup$

We have

$$\bar{z} - z^2 = i(\bar{z} + z^2)$$

and multiplying by $i$

$$-\bar{z} - z^2 = i(\bar{z} -z^2)$$

then summing

$$z^2=-i\bar z\iff r^2e^{i2\theta}=re^{i\left(-\theta+\frac 3 2\pi\right)}$$

from which we can conclude that $r=0$ or $r=1$ with

$$2\theta=-\theta+\frac 3 2\pi+2k\pi \iff \theta =\frac \pi 2 +\frac 2 3 k\pi$$

$\endgroup$
1
$\begingroup$

Let $z=x+iy$

$$z^2 =\bar z\cdot\dfrac{1-i}{1+i}=\cdots=-i\cdot\bar z$$

$$\implies x^2-y^2+i(2xy)=-i(x-iy)=-y-ix$$

Equating the imaginary parts, $-x=2xy\iff x(2y+1)=0\ \ \ \ (1)$

Equating the real parts, $x^2-y^2=-y\ \ \ \ (2)$

From $(1),$

either $x=0, $ using $(2), 0^2-y^2=-y\implies y=?$

or $2y+1=0, $ using $(2), x^2=y^2-y=?\implies x=?$

$\endgroup$
2
  • $\begingroup$ It is cleaner not to divide by $\overline z$ at the begining and then re-multiply by it. Better keep the equation like this : $$z^2=\frac{1-i}{1+i}\overline z=-i\overline z.$$ $\endgroup$ Commented Aug 31, 2022 at 7:04
  • $\begingroup$ @AnneBauval, thanks for your observation $\endgroup$ Commented Aug 31, 2022 at 7:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .