I am trying to connect $a,b,n$ such that $$\Im\left ((a+bi)^n \right )=0, a,b \in \mathbb{R}, n \in \mathbb{N^*} \mathrm{or } \; \mathbb{R^*}$$

What I tried was write $(a+bi)^n$ as $\sqrt{a^2+b^2} e^{n i \theta}$ where

$$\theta= \begin{cases} \arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a,b>0 \\ \pi- \arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a<0, b>0 \\-\arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a>0, b<0 \\-\left[\pi-\arctan\left(\frac{b}{a}\right)\right] \quad \text{if} \quad a<0, b<0 \\ \quad \\ 0 \quad \text{if} \quad a>0, b=0 \\ \quad \\ \frac{\pi}{2} \quad \text{if} \quad a=0, b>0\ \\ \quad \\ \pi \quad \text{if} \quad a<0, b=0 \\ \quad \\ -\frac{\pi}{2} \quad \text{if} \quad a=0, b<0 \\ \quad \\ \text{undefined} \quad \text{if} \quad a=b=0 \end{cases} \quad ,$$

Then it must hold that $\frac{n \theta}{\pi} \in \mathbb{Z}$ if we want $\Im\left( (a+bi)^n\right )=0$

This is where I'm stuck as my number theory(that's what I suspect is needed to connect $a,b,n$) knowledge is close to non-existent.

The "or" up there is an extension of the problem.

  • $\begingroup$ Sorry, what do you mean by 'connect'? $\endgroup$ – Prometheus Sep 17 '14 at 11:43
  • $\begingroup$ Also that definition of $\theta$ is horrifying. You really don't need it. Just saying let $a+bi = re^{i\theta}$, everyone will know what you mean. $\endgroup$ – Prometheus Sep 17 '14 at 11:46
  • $\begingroup$ where $r=\sqrt{x^2+y^2}$. $\endgroup$ – Prometheus Sep 17 '14 at 11:46
  • 1
    $\begingroup$ Well, if you're going to use only integer powers, you can't do it to just any complex number. The number $\theta/\pi$ has to be rational. If it's equal to $a/b$ where $gcd(a,b)=1$, raise to the power of $b$. $\endgroup$ – Prometheus Sep 17 '14 at 11:48
  • 1
    $\begingroup$ Alright well, you want $n\theta /\pi$ to be an integer, then let $n = \pi/\theta$. $\endgroup$ – Prometheus Sep 17 '14 at 11:54

$$Im(e^{in\theta})=\sin(n\theta)=0 \iff n\theta=k\pi$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.