How do you solve $z^4 = 2(1+i\sqrt{3})$

Question

Solve $z^4 = 2(1+i\sqrt{3})$ in the form $r(\cos\alpha+i\sin\alpha)$ where $r>0$ and $0\le\alpha<2\pi$

I know you have to find $\arctan(\frac{\sqrt{3}}{1})=\frac{\pi}{3}$ and that is $\alpha$? I am not really sure how to go about doing this.

hint: $1+i\sqrt{3}=\sqrt{1^2+\sqrt{3}^2}\left(\frac 12+i\frac{\sqrt{3}}2\right)=2,e^{i\pi/3}$ — Raymond Manzoni, Apr 27 '13 at 10:59
One huge contributor to the community recently answered a question that described the process of solving this kind of problems. I can't remember who it was. Hopefully someone will know what I'm talking about and provide the link to that answer. Despite this I'm sure there are other similar questions here and this should probably be marked as a duplicate. — Git Gud, Apr 27 '13 at 10:59
The $\sqrt{2}$ is right but you have to consider the different roots of $e^{i\pi/3}$ (the four roots $e^{(i\pi/3+2k\pi i)/4}$ !) — Raymond Manzoni, Apr 27 '13 at 11:07
Replace for example $k$ by $-1,0,1,2$ to get $e^{i\pi/{12}-\pi i/2},\ e^{i\pi/{12}},\cdots$ (that you may write in trigonometric form if you prefer) — Raymond Manzoni, Apr 27 '13 at 11:14
$z = \sqrt2(\frac{1}{2}+i\frac{\sqrt{3}}{2})$ = $cos\alpha + isin\alpha = e^{i\alpha}$ — maxmitch, Apr 27 '13 at 11:16
Well that was wrong earlier and remains wrong :-). $z=\sqrt{2}\left(\cos(\pi/12)+i\sin(\pi/12)\right)$ and so on (the angle has to be divided by $4$, see DonAntonio's answer) — Raymond Manzoni, Apr 27 '13 at 11:18
@maxmitch Do you understand the goal of this question? Meaning, do you understand the two ways to represent $z$? One is the "cartesian" coordinates for $z$ and the other is "trigonometric". Euler's identity helps you perform this transformation because taking the 4th root of $z$ is ugly when you don't have it in the "complex exponential" form (the $e^{i\alpha}$ stuff). — roliu, Apr 27 '13 at 11:22
I understand that you are supposed to find the modulus and the argument. But I don't understand the steps to find the modulus. And I don't know what to do once you have found them both... — maxmitch, Apr 27 '13 at 11:24
Try the easier problem where we want to transform $z = 2(1 + i\sqrt{3})$ into its trigonometric form instead of $z^4 = 2(1 + i\sqrt{3})$. If you can do that then you can do one of the main steps for your original problem. You should be able to do the easier problem by just rereading the definitions carefully and drawing it on paper. — roliu, Apr 27 '13 at 11:26
You got $z^4=4,e^{i\pi/3+2k\pi i}$ (since adding $2\pi$ should change nothing). From this deduce that $z=\sqrt{2}e^{(i\pi/3+2k\pi i)/4}=\sqrt{2}e^{i\pi/{12}+k\pi i/2}=\sqrt{2}(\cos(\pi/12+k\pi/2)+i\sin(\pi/12+k\pi/2))$ — Raymond Manzoni, Apr 27 '13 at 11:27
Where did he claim to understand the equivalence: $z^4 = 4e^{i\pi/3}$ ? I really think he needs to take this slower and that the existing answers aren't of much help to him. But perhaps it's the wrong forum too. — roliu, Apr 27 '13 at 11:30
@roliu (if you are talking to me) : he didn't say that he didn't know them either (and he speaks of modulus and argument). That's why there is a discussion... (sorry I have to go...) — Raymond Manzoni, Apr 27 '13 at 11:33
@RaymondManzoni He knows that the exponential form for a complex number is composed of two parts the "modulus" and the "argument". He doesn't seem to know how those are defined (hence why I suggested he reread the definitions carefully). How would he understand that equivalence without that knowledge? And you are right about him not claiming that he didn't know it. But uh I haven't claimed to not know the proof for Fermat's Last Theorem. Valid to suggest that I do know it? Whatever, separate argument. — roliu, Apr 27 '13 at 11:37
Related: http://math.stackexchange.com/questions/192742/how-to-solve-x3-1/192743#192743 — lab bhattacharjee, Apr 27 '13 at 12:11
This detailed explanation may help you too 'Trigonometric Form of Complex Numbers'. Once you get a trigonometric form (ignoring the modulus $R$) $\cos(\alpha)+i\sin(\alpha)$ you may rewrite it as $e^{i\alpha}$ this is Euler's formula (you may consider this exponential form a 'shortcut' if you want with the advantage that the usual formulas for exponentials apply). — Raymond Manzoni, Apr 27 '13 at 12:25
See here. Do not forget to up vote the answers that you benefit from them. — Mhenni Benghorbal, Apr 27 '13 at 13:58

score 2 · Answer 1 · answered Apr 27 '13 at 11:16

Hints:

$$(1)\;\;\;w:=2+2\sqrt 3\,i\implies |w|=\sqrt{4+12}=4\;,\;\;\arg w=\arctan\frac{2\sqrt 3}{2}=\frac{\pi}{3}\;\implies$$

$$(2)\;\;z^4=\left(re^{i\phi}\right)^4=w=4e^{\frac{\pi i}{3}+2k\pi i}\implies z_k=4^{1/4}e^{\frac{\pi i}{12}(1+2k)}\;,\;\;k=0,1,2,3\; (\text{why only these?})$$

score 1 · Answer 2 · answered Apr 27 '13 at 11:05

1

$$cos(\alpha)+isin(\alpha)=e^{i\alpha}$$ $$z^{4}=re^{i\alpha}\implies z=(r)^{1/4}e^{i\alpha/4}=r^{1/4}[cos(\alpha /4)+isin(\alpha/4)]$$

Here r=4 and $\alpha=\dfrac{\pi}{3}$

answered Apr 27 '13 at 11:05

Shaswata

5,068

How do you solve $z^4 = 2(1+i\sqrt{3})$

2 Answers2