Can some one explain to me what is going on here - power of complex number

Question

So here is the question and the work to solve it, but I have no idea how one knows to do the first step or what the first step is...

$$ \begin{align} (6-i\sqrt{12})^{12} &= \left[\sqrt{48}\left(\cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)\right)\right]^{12}\\ &= (\sqrt{48})^{12} \left[\cos\left(\frac{12\pi}{6}\right) - i\sin\left(\frac{12\pi}{6}\right)\right]\\ &=48^6 \end{align} $$

Yeah, I didn't know how to format it correctly so just did what I could. Thanks. — user141750, Jun 11 '14 at 18:57
For a proper tutorial about formatting in TeX, I suggest you to refer to: MathJax basic tutorial and quick reference — jojeck, Jun 11 '14 at 19:03
For what it's worth, I would rewrite $6 - i \sqrt{12}$ as $6 - 2 \sqrt{-3}$. — James47, Jun 11 '14 at 21:12
Is there any particular reason why the OP has to go through the sines and cosines? If the point here is to go from $(6 - \sqrt{-12})^2$ to $48^6$ it would make much more sense to compute squares and cubes as algebraic integers. Maybe that's what James is getting at, though perhaps 141750 can (for now) do without knowing the fact that $\mathbb{Z}[\sqrt{-12}]$ is a subdomain of $\mathbb{Z}[\omega]$. — Robert Soupe, Jun 12 '14 at 05:01

score 3 · Accepted Answer · answered Jun 11 '14 at 18:53

3

Step 1 Write your number in polar coordinates, i.e. $z=r(\cos t+i\sin t)$.

Step 2 Use De Moivre's theorem, that $(\cos t+i\sin t)^n=\cos nt+i\sin nt$.

answered Jun 11 '14 at 18:53

Pedro

122,002

It looks as if he's already done those steps. – Michael Hardy Jun 11 '14 at 19:20
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@MichaelHardy It looks to me as he's reading them off somewhere and isn't understanding them, hence the explanation. – Pedro Jun 11 '14 at 19:39

score 3 · Answer 2 · answered Jun 11 '14 at 20:09

Some of the other posts have given general theory let be just point out how to analyse the question. We have $6-i\sqrt{12}$ we want to write it as $\sin$ and $\cos$ so we can use DeMovire. We need first to find its length,

$$\sqrt{6^2+\sqrt{12}^2} =\sqrt{48}=4\sqrt{3}$$

the normalized number is $$\frac{6-i\sqrt{12}}{4\sqrt{3}}=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$

Thus we need $\theta$ such that

$$\cos \theta =\frac{\sqrt{3}}{2}$$ and $$\sin \theta =-\frac{1}{2}$$ from our knowledge of trig and the unit circle we see that $\theta=\frac{\pi}{6}$. The rest you know...

score 1 · Answer 3 · answered Jun 11 '14 at 19:17

It’s not clear to me what you want for an explanation, but let me try. By direct computation, you can show this, if $r,r'$ are positive reals and $\theta, \theta'$ are angles (or real numbers representing radian measure, if you like): $$ [r(\cos\theta+i\sin\theta)]\cdot[r'(\cos\theta'+i\sin\theta')]=rr'\bigr(\cos(\theta+\theta')+i\sin(\theta+\theta')\bigr)\,, $$ using the addition formulas for cosine and sine, namely $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\sin(A+B)=\sin A\cos B+\cos A\sin B$. Geometrically, this means that when you multiply two complex numbers of distance $r$ and $r'$ respectively, the new distance is $rr'$, while if their angles (from the positive real axis) are $\theta$ and $\theta'$ respectively, the product is found in the direction $\theta+\theta'$. Distances multiply, angles add.

All follows from this.

Can some one explain to me what is going on here - power of complex number

3 Answers3