# How can I find modulus and argument of $z=(1+i)^n$ [closed]

The complex numbers are such that : $z=(1+i)^n$. Find the modulus and argument of complex-numbers.

• Begin by finding modulus and agument of $1+i$. Nov 4, 2017 at 21:24
• and after?..... Nov 4, 2017 at 21:30
• I'm voting to close this question as off-topic because the OP has shown no personal effort and appears to be merely asking someone to work their homework problem. Nov 4, 2017 at 21:42

Either by DeMoivre's formula or Euler's identity, it can be shown that for complex $z$ and integer $n$, $|z^n| = |z|^n$. So $|(1+i)^n| = |1+i|^n = \sqrt 2^n$. With the same idea, write $z$ in the trig form, prove that $\mathrm{cis} a \cdot \mathrm{cis} b = \mathrm{cis} (a+b)$ so it follows that $z^n = |z|^n \mathrm{cis} (n\theta)$. Try to find $\mathrm{arg} (1+i)$ as an exercise.

• Prove that if $a, b \in \Bbb C$, then $|ab| = |a||b|$. Then use induction. Nov 4, 2017 at 21:40
• thank you and what about argument ? Nov 4, 2017 at 21:42
• Oh my, sorry. I'll complete it. Nov 4, 2017 at 21:45
• okay i wait you thank you Nov 4, 2017 at 21:47

Hint Write $1+i$ in trigonometric form.

• how ,,?......... Nov 4, 2017 at 21:24
• @user499303 Your teacher told you what the modulus is. Simply use the definition. Nov 4, 2017 at 21:25
• i know what is the modulus but i can't find this only :) Nov 4, 2017 at 21:26
• and what about ^n Nov 4, 2017 at 21:37

You can use something like that:
$$\forall z\in \mathbb{C}: z = |z|(\cos(\alpha)+i\sin(\alpha))$$ Where $|z| = \sqrt{a^2+b^2}$ if $z = a + bi; a, b \in \mathbb{R}$
Then you can use this" de Moivre formula.

Thus we get: $$z^n = |z|^n(\cos(n\alpha)+i\sin(n\alpha))$$

• z=(1+i)^n ?,,,, Nov 4, 2017 at 21:36
• I have edited my post so you can use it to calculate $z^n$ Nov 4, 2017 at 21:39