Compute complex number $(1 + i)^{14}$

Question

Compute the complex numbers, writing your answers in the standard form $a + bi$

a) $(1 + i)^{14}$

So I am trying to use polar coordinates and De Moirre's formula to compute this but I ran into a problem. I started out by writing it in the form:

$(r cis$$\theta$$)^{14}$ = $r^{14}$cis($14\theta$)

However, when solving for $r = \sqrt{x^2 + y^2}$ that would be $1 + i^2$ which would be $1 + (-1) = 0$. Unless I am missing something which would make my whole computation $0$.

I don't believe this is the case so I am just wondering if I am thinking about this technique incorrectly or if there is another way to go about this.

Thank you.

Update: With the help I've been able to come to an answer.

So I figured out 1 + i = $\sqrt{2}$cis($\frac{\pi}{4}$)

$(1 + i)^{14}$ = $\sqrt{2}^{14}$($\frac{\pi}{4}$ x 14) = $2^7$($\frac{\pi}{4}$ x 14)

$2^7$ cis($\frac{14\pi}{4}$) = $2^7$ (0 - 1i)

= 0 - 128i

If anyone can verify what I did, I would be appreciative.

Answer 1

Your number $y$ is not $i$. It is $1$, so that $x+yi=1+i$. Therefore, $\sqrt{x^2+y^2}=\sqrt2$ and$$1+i=\sqrt2\left(\cos\left(\frac\pi4\right)+\sin\left(\frac\pi4\right)i\right).$$

Answer 2

You need to write $1+i$ in polar. You have $x+yi = 1+i$, what are the values of $x$ and $y$? Find $r = \sqrt{x^2+y^2}$.

Answer 3

Hint: $\require{cancel}\;(1+i)^2 = \bcancel{1} + 2i + \bcancel{i^2} = 2i\,$.