Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm here because I can't finish this problem, that comes from a Russian book:

Calculate $z^{40}$ where $z = \dfrac{1+i\sqrt{3}}{1-i}$

Here $i=\sqrt{-1}$. All I know right now is I need to use the Moivre's formula $$\rho^n \left( \cos \varphi + i \sin \varphi \right)^n = \rho^n \left[ \cos (n\varphi) + i \sin (n\varphi) \right]$$

to get the answer of this.

First of all, I simplified $z$ using Algebra, and I got this:

$$z = \dfrac{1-\sqrt{3}}{2} + i \left[ \dfrac{1+\sqrt{3}}{2} \right]$$

Then, with that expression I got the module $|z| = \sqrt{x^2 + y^2}$, and its main argument $\text{arg}(z) = \tan^{-1} \left( \dfrac{y}{x} \right)$.

I didn't have problems with $|z| = \sqrt{2}$, but the trouble begins when I try to get $\text{arg}(z)$. Here is what I've done so far:

$$\alpha = \text{arg}(z) = \tan^{-1} \left[ \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right]$$

I thought there's little to do with that inverse tangent. So, I tried to use it as is, to get the power using the Moivre's formula.

$$z^{40} = 2^{20} \left[ \cos{40 \alpha} + i \sin{40 \alpha} \right]$$

As you can see, the problem is to reduce a expression like: $\cos{ \left[ 40 \tan^{-1} \left( \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right) \right] }$.

And the book says the answer is just $-2^{19} \left( 1+i\sqrt{3} \right)$.

I don't know if I'm wrong with the steps I followed, or if I can reduce those kind of expressions. I'll appreciate any help from you people :)

Thanks in advance!

share|cite|improve this question
All you need to work out is what angle that arctangent represents; the rest follows straightly. – Semiclassical Aug 17 '14 at 23:03
Those russian texts are always fascinating by the manner they carefully choose the straightforward but nevertheless not so evident to work out problems :) – Math Gems Aug 17 '14 at 23:04
Yes, but the problem is I can't use a calculator to get that angle. Is there a way to do so w/o it? – TX286 Aug 17 '14 at 23:05
Hrm, you probably can but it gets a bit hairy. There's a better hint to start from, which I'll give below...actually, scratch that. My hint is superfluous given Troy's answer below. – Semiclassical Aug 17 '14 at 23:09
You're right @MathGems :) Ok, I'll check those answers ;) – TX286 Aug 17 '14 at 23:13
up vote 4 down vote accepted

Oh my. We have: $$1+i\sqrt{3} = 2\exp\left(i\arctan\sqrt{3}\right)=2\exp\frac{\pi i}{3}$$ $$\frac{1}{1-i} = \frac{1}{2}(1+i) = \frac{1}{\sqrt{2}}\exp\frac{\pi i}{4},$$ hence: $$z=\frac{1+i\sqrt 3}{1-i} = \sqrt{2}\exp\frac{7\pi i}{12},$$ so: $$ z^{40} = 2^{20}\exp\frac{70\pi i}{3}=2^{20}\exp\frac{4\pi i}{3}=-2^{20}\exp\frac{\pi i}{3}=-2^{19}(1+i\sqrt{3}).$$ As an alternative way, if you set $a=1+i\sqrt{3}$ and $b=\frac{1}{1-i}$ you have: $$ a^3 = -2^3,\qquad b^4 = -2^{-2}, $$ hence: $$ z^{40} = a^{40} b^{40} = a(a^3)^{13} (b^4)^{10} = a\cdot(-2^{39})\cdot(2^{-20}) = -2^{19}(1+i\sqrt{3}).$$

share|cite|improve this answer
@jackdaurizio, thank you! You helped me a lot! Just one thing: I think pi/4 in the second exp would be -pi/4, am I right? – TX286 Aug 17 '14 at 23:51
No, it should be right, $(1+i)$ has a positive imaginary part. – Jack D'Aurizio Aug 17 '14 at 23:54
Oh, I see. You're right! Thank you again! ;) – TX286 Aug 17 '14 at 23:58

$$ z=\sqrt 2\frac{\frac{1+\sqrt 3i}{2}}{\frac{\sqrt 2-\sqrt 2i}{2}}=\sqrt 2 e^{i(\pi/3-(-\pi/4))} $$ and therefore: $$ z^{40}=2^{20}e^{i2\pi/3}=-2^{19}(1+i\sqrt3) $$

share|cite|improve this answer

To tackle the arctangent head-on, we make use of an identity derived from the tangent angle addition formula:

$$\tan\left(\frac\pi4+x\right)=\frac{\tan\frac\pi4+\tan x}{1-\tan\frac\pi4 \tan x}=\frac{1+\tan x}{1-\tan x}$$ since $\tan \dfrac\pi4=1$. Recalling that $\tan\dfrac\pi3=\sqrt{3}$, then we immediately have $$\alpha=\tan^{-1}\left(\frac{1+\sqrt{3}}{1-\sqrt{3}}\right)=\tan^{-1}\left(\frac{1+\tan\frac\pi3}{1-\tan\frac\pi3}\right)=\tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\right]=-\frac{5\pi}{12}$$ since $\tan^{-1}$ takes values in $(-\pi/2,\pi/2]$.

From here, things are clear: $$\cos(40\alpha)+i\sin(40\alpha) =\cos\left(-\frac{2\pi}{3}-16\pi \right)+i\sin\left(-\frac{2\pi}{3}-16\pi \right)=-\frac{1+i\sqrt{3}}{2}$$ Multiplying through by the modulus $2^20$ gives the desired result.

share|cite|improve this answer
Note that this is not the 'clever' approach; it is certainly better to recognize that $z$ is a quotient of simpler complex numbers. But the $\tan\left(\frac{\pi}{4}+x\right)$ identity given above is a nice addition to one's toolkit. – Semiclassical Aug 18 '14 at 0:12
Thank you very much @semiclassical, it's great to know this! ;) – TX286 Aug 18 '14 at 0:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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