1

How do I rewrite this to 'a + ib' form? The power of 30 is troubling me.

$ (\sqrt{3}+ i)^{30} $

javanewbie
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3 Answers3

5

Hint. Since $|\sqrt{3}+i|=2$, we have that $$(\sqrt{3}+ i)^{30}=2^{30}\left(\frac{\sqrt{3}}{2}+ \frac{i}{2}\right)^{30}=2^{30}\left(\cos(\pi/6)+ i\sin(\pi/6)\right)^{30}.$$ Then use De Moivre's Formula.

Robert Z
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  • Hmm, but how did you figure out you had to divide by 2? – javanewbie Nov 15 '16 at 17:10
  • @javanewbie Because $|\sqrt{3}+i|=2$. – Robert Z Nov 15 '16 at 17:12
  • I am just learning about complex numbers so excuse my ignorance, but I still don't quite understand. Why it was specifically the number 2 I understand now, but why divide it by anything? And how does it translate into the last part? – javanewbie Nov 15 '16 at 17:17
  • In order to use De Moivre's Formula you should have a complex number of modulus one that is $\frac{\sqrt{3}}{2}+ \frac{i}{2}$. Then any complex number of modulus one can be written as $\cos(\theta) +i\sin(\theta)$ for some angle $\theta$ (the argument). – Robert Z Nov 15 '16 at 17:21
  • See https://en.wikipedia.org/wiki/Complex_number for more details – Robert Z Nov 15 '16 at 17:23
  • Yes but not sure if I'm just not understanding out of stupidity, heh. But what do you mean by 'number of modulus one' (English isn't exactly my first language so the terminology varies a bit) – javanewbie Nov 15 '16 at 17:29
  • @javanewbie "Modulus" is the same of "absolute value". The absolute value of $z=x+iy$ is $|z|=\sqrt{x^2+y^2}$. See https://en.wikipedia.org/wiki/Complex_number#Absolute_value_and_argument – Robert Z Nov 15 '16 at 17:32
5

"By hand":

$$(\sqrt3+i)^2=3-1+i2\sqrt3=2(1+i\sqrt3)$$

$$(\sqrt3+i)^3=2(1+i\sqrt3)(\sqrt3+i)=8i.$$

Then

$$(\sqrt3+i)^{30}=(8i)^{10}=(-64)^5=-1073741824.$$

4

Hint. Use exponential/polar form $$\sqrt{3}+i = 2e^{\frac{\pi}{6}i}.$$

Eff
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