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I understand how to write $(1+i)$ in polar form, but how do I use it to compute $(1+i)^{10}$?

Thanks for the help!

Rajat
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Nick
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    Multiply the angle by 10 and the raise the radius to the 10th power. – Gregory Grant Dec 01 '15 at 19:15
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    Hi and welcome to the site, it is advised to learn some typesetting - at least once you have been here for a while. Also important to show any efforts you have tried to solve the problem. Otherwise if the question looks too much like a homework question you are likely to get negative feedback if not showing some effort.

    A hint you can use is that the argument of the complex number multiplies with the exponent and the magnitude is exponentiated with the exponent.

     – mathreadler Dec 01 '15 at 19:15
  • Alright, I appreciate the feedback, thank you very much! – Nick Dec 01 '15 at 19:17

3 Answers3

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$$1+i=|1+i|e^{\arg(1+i)i}=$$ $$\sqrt{\Re(1+i)^2+\Im(1+i)^2}e^{\arctan\left(\frac{\Im(1+i)}{\Re(1+i)}\right)i}=$$ $$\sqrt{1^2+1^2}e^{\arctan\left(\frac{1}{1}\right)i}=$$ $$\sqrt{2}e^{\arctan\left(1\right)i}=$$ $$\sqrt{2}e^{\frac{\pi}{4}i}$$

So:

$$\left(1+i\right)^{10}=$$ $$\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^{10}=$$ $$\left(\sqrt{2}\right)^{10}\cdot e^{10\cdot\frac{\pi}{4}i}=$$ $$\sqrt{2^{10}}e^{\frac{10\pi}{4}i}=$$ $$\sqrt{1024}e^{\frac{5\pi}{2}i}=$$ $$32e^{\left(2\pi+\frac{\pi}{2}\right)i}=$$ $$32e^{\frac{\pi}{2}i}=$$ $$32i$$

Jan Eerland
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Polar form is $r(\cos\theta + i\sin\theta)$. The $10$th power is $$ \Big(r(\cos\theta + i\sin\theta)\Big)^{10} = r^{10}(\cos(10\,\theta) + i\sin(10\,\theta)). $$

The fact that $$ (\cos\theta + i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta) $$ appears to be the part you missed. It is sometimes called de Moivre's formula, after Abraham de Moivre, an 18th-century mathematician. I was puzzled by its being called a theorem when I first saw it in a secondary-school textbook, because I had been taught some years earlier that this is the definition of multiplication of complex numbers.

Michael Hardy
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Hint: $a+bi=re^{i\theta}=r(\cos \theta +i \sin \theta)$ where $r=\sqrt{a^2+b^2}$ and $\tan \theta = \frac{b}{a}$.

$(a+bi)^k=r^k e^{i k \theta}=r^k(\cos (k \theta) +i \sin (k\theta))$

Rajat
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