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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!

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    $\begingroup$ All you need to work out is what angle that arctangent represents; the rest follows straightly. $\endgroup$ – Semiclassical Aug 17 '14 at 23:03
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    $\begingroup$ Those russian texts are always fascinating by the manner they carefully choose the straightforward but nevertheless not so evident to work out problems :) $\endgroup$ – Math Gems Aug 17 '14 at 23:04
  • $\begingroup$ 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? $\endgroup$ – user170247 Aug 17 '14 at 23:05
  • $\begingroup$ 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. $\endgroup$ – Semiclassical Aug 17 '14 at 23:09
  • $\begingroup$ You're right @MathGems :) Ok, I'll check those answers ;) $\endgroup$ – user170247 Aug 17 '14 at 23:13
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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: $$ z^{40} = a^{40} b^{40} = a(a^3)^{13} (b^4)^{10} = a\cdot(-2^{39})\cdot(2^{-20}) = -2^{19}(1+i\sqrt{3}).$$

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  • $\begingroup$ @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? $\endgroup$ – user170247 Aug 17 '14 at 23:51
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    $\begingroup$ No, it should be right, $(1+i)$ has a positive imaginary part. $\endgroup$ – Jack D'Aurizio Aug 17 '14 at 23:54
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    $\begingroup$ Oh, I see. You're right! Thank you again! ;) $\endgroup$ – user170247 Aug 17 '14 at 23:58
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$$ z=\sqrt 2\frac{\frac{1+\sqrt 3i}{2}}{\frac{\sqrt 2-\sqrt 2i}{2}}=\sqrt 2 e^{i(\pi/3-(-\pi/4))} $$ and therefore: $$ z^{40}=2^{20}e^{i2\pi/3}=-2^{19}(1+i\sqrt3) $$

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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.

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    $\begingroup$ 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. $\endgroup$ – Semiclassical Aug 18 '14 at 0:12
  • $\begingroup$ Thank you very much @semiclassical, it's great to know this! ;) $\endgroup$ – user170247 Aug 18 '14 at 0:31

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