0
$\begingroup$

I am currently prepping for next year's math course. Currently, I am practicing complex numbers and I have come across something I don't understand.

The problem is how to write the following complex number in rectangular form: $(1+i)^{13}$

I know the argument(z) is $\dfrac{(13 \cdot \pi)}{4}$

but when trying to find modulus, the solution says it is $\sqrt{2}^{13}$. I follow this so far but I don't get what the rule is for the next step, which is $2^6 \cdot \sqrt(2)$. What is happening here?

Any help is very much appreciated!

$\endgroup$
1
  • $\begingroup$ Welcome to MSE, please, learn how to use mathjax to type your questions $\endgroup$ – jjagmath Dec 28 '20 at 10:38
2
$\begingroup$

Hint: $x^{13}=x^{12} x = (x^2)^6 x$

$\endgroup$
0
$\begingroup$

The absolute value is$$\left|(1+i)^{13}\right|=|1+i|^{13}=\sqrt2^{13}$$and\begin{align}\sqrt2^{13}&=\sqrt2^{12}\times\sqrt2\\&=\left(2^{1/2}\right)^{12}\times\sqrt2\\&=2^6\sqrt2.\end{align}

$\endgroup$
1
  • $\begingroup$ @Forester I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 28 '20 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.