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I'm having trouble with complex numbers. For example, I need to find all the solutions to

$$ z^4 = i \bar z ^3$$

My attempt was $$z= |z|\ \left(\cos(\alpha)+i\ \sin(\alpha)\right) $$ $$ z^4 = i \bar z ^3$$ $$z = i \left({\frac {\bar z}{z}}\right) ^3 $$ $$z = i \left({\frac {\bar z}{|z|}}\right) ^6 = i\left(\cos(-6\ \alpha)+i\ \sin(-6\ \alpha)\right) = i\ \cos(-6\ \alpha)- \sin(-6\ \alpha)$$ $$|z|\ \left(\cos(\alpha)+i\ \sin(\alpha)\right) = i\ \cos(-6\ \alpha)- \sin(-6\ \alpha)$$

From where you get $$|z|\ \cos(\alpha)= - \sin(-6\ \alpha)$$ $$ |z| \sin(\alpha)= \cos(-6\ \alpha)$$

From that, I got: $$\cos (7\alpha) = 0$$ Then $$ 7\alpha= \frac {\pi}{2} + k\pi \qquad k\in \mathrm {Z} \\ \alpha= \frac {\pi}{14} + \frac{\pi}{7}k$$

With this I get $|z| =1$

Is it okay? And if it is, isn't there any easier way to do it? I think I'm overcomplicating things...

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It becomes easier using Euler's formula $e^{i\alpha}=\cos\alpha+i\sin\alpha$. The equation simplifies to $re^{i7\alpha}=i=e^{i\pi}$, implying $r=1$ and $7\alpha = \pi \pmod{2\pi}$. – Jonathan Y. Aug 18 '13 at 23:37
We haven't seen Euler's formula so I don't think I can use – milo Aug 18 '13 at 23:39
Well, one can utilize trig identities to show $\frac{\operatorname{cis}(\alpha)}{\operatorname{cis}(\beta)}=\operatorname{cis}‌​(\alpha-\beta)$, to much the same effect. – Jonathan Y. Aug 18 '13 at 23:43
up vote 7 down vote accepted

You got to almost exactly the right place, It is easier if we hold off introducing sines and cosines.

There is the obvious solution $z=0$. We now look for the non-zero solutions.

For non-zero $z$, taking norms, we find that the norm of $z$ must be $1$.

If we multiply both sides by $z^3$, we get the equivalent equation $z^7=i$. Now it's essentially over. For $i=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}$. Use De Moivre's Formula to write down the $7$-th roots of $i$.

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Much simpler, this is probably what they were aiming at. Thanks! – milo Aug 18 '13 at 23:55
You are welcome. It is not all that different from yours. You lost the $0$ root. And you have $\frac{k\pi}{7}$ instead of $\frac{2k\pi}{7}$. – André Nicolas Aug 19 '13 at 0:01

When you start to involve powers of complex numbers, working with the exponential form tends to be easier:

$$z=|z|e^{i\phi}$$ The equation is:


From here, we have the soolution $z=0$, if $z\neq 0$ then we can divide by the module of $z$, and changing $i=e^{i\pi/2}$, we have:

$$|z|e^{7i\phi}=e^{i\frac{\pi}{2}+2k\pi i}\qquad k\in\mathbb{Z}$$

We conclude that $|z|=1$ and


The solutions will start repeating when $\phi$ goes all the way around ($2k\pi/7=2\pi$), which means that you will get the seven different solutions for $k=0,1,2,3,4,5,6$.

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Let $z=re^{i\theta}.$ Clearly, $z=0$ is solution so let us assume that $z\neq0.$ So, from the given equation, \begin{align} & &r^4e^{4i\theta}=ir^3e^{-3i\theta}\\ &\implies& re^{7i\theta}=i\\ &\implies & r\cos(7\theta)+ir\sin(7\theta)=i\\ &\implies & r\cos(7\theta)=0 \implies \cos(7\theta)=0\implies 7\theta=(2n+1)\pi/2\\ &\text{and}& r\sin(7\theta)=1 \implies r=1 \end{align} So $z=e^{i\theta}; \theta=(2n+1)\pi/14.$

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Note that $i=e^{i\pi/2}$ and write $z=re^{i\theta}$. Thus $z^4=i\bar z^3$ becomes $r^4e^{i4\theta}=r^3e^{i(\pi/2-3\theta)}$, from which $r=1$ and $4\theta=\pi/2-3\theta+2k\pi$ that yelds to $\theta=\frac17(\pi/2+2k\pi),\;\;k=0,\dots,6$.

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