# Find the real part in a complex number [closed]

Today I was faced with the following question:

Find the real part in the complex number $z = (1 + i)^{12}$.

How should I proceed on that endeavor?

• what have you tried? Do you know how to write complex numbers in polar form? Do you know how to exponentiate complex numbers? Jan 8, 2017 at 5:17
• What is $(1+i)^2$? Then what is $((1+i)^2)^{50})$? Jan 8, 2017 at 5:21

$z=((1+i)^2)^6$

$=(1-1+2i)^6$

$=(2i)^6$

$=2^6.i^6$

$=-64$

• This is what I was getting at too. @oqrxke what do you think you would do if it were $(1+i)^{17}$? Jan 8, 2017 at 5:29
• $(1+i)^{17}=(1+i)^{16}*(1+i)=-64*(1+i)=-64-64i$ Therefore, the real part of $(1+i)^{17}$ would still be -64 Jan 8, 2017 at 6:46
• That's strange, no? Jan 8, 2017 at 16:09

Converting into mod-arg form (polar form); $1+i=\sqrt{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$.

Then utilise De-Moivre's Theorem

Consider that: $$z=|z|e^{Arg(z)}$$ Then it is $$z^{n}=|z|^{n}e^{nArg(z)}$$

We see that $$\mbox{ the modulus } |z|=\sqrt{2}, \qquad \arg z = {\pi\over4}.$$ When we raise $z$ to the $12$th power, the modulus is raised to the $12$th power while the argument is multiplied by $12$, so $$|z^{12}|=(\sqrt{2})^{12}=2^6 = 64, \qquad \arg z = 12\times{\pi\over4} = 3\pi.$$ Therefore $z^{12}$ is located to the left from the origin, at the distance 64 from the origin. So $$z^{12} = -64.$$ So $z^{12}$ is on the real axis; thus the real part of $z^{12}$ is $$\mbox{Re } z^{12} = -64.$$

HINT : represent $\displaystyle (1+i)^{12} = (\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})))^{12}$