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

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    $\begingroup$ Begin by finding modulus and agument of $1+i$. $\endgroup$
    – GEdgar
    Nov 4, 2017 at 21:24
  • $\begingroup$ and after?..... $\endgroup$ Nov 4, 2017 at 21:30
  • 1
    $\begingroup$ Unfortunately, I feel I had to flag this. This isn’t a site that will solve your homework for you, and some effort or examples of your own thoughts must be shown. That way, we can help you—and others—the best we can. For more information, please read how to ask a good question. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! $\endgroup$ Nov 4, 2017 at 21:38
  • 4
    $\begingroup$ 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. $\endgroup$ Nov 4, 2017 at 21:42

3 Answers 3


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.

  • $\begingroup$ Prove that if $a, b \in \Bbb C$, then $|ab| = |a||b|$. Then use induction. $\endgroup$ Nov 4, 2017 at 21:40
  • $\begingroup$ thank you and what about argument ? $\endgroup$ Nov 4, 2017 at 21:42
  • $\begingroup$ Oh my, sorry. I'll complete it. $\endgroup$ Nov 4, 2017 at 21:45
  • $\begingroup$ okay i wait you thank you $\endgroup$ Nov 4, 2017 at 21:47
  • $\begingroup$ I HAVE other number please help me $\endgroup$ Nov 4, 2017 at 22:58

Hint Write $1+i$ in trigonometric form.

  • $\begingroup$ how ,,?......... $\endgroup$ Nov 4, 2017 at 21:24
  • $\begingroup$ @user499303 Your teacher told you what the modulus is. Simply use the definition. $\endgroup$ Nov 4, 2017 at 21:25
  • $\begingroup$ i know what is the modulus but i can't find this only :) $\endgroup$ Nov 4, 2017 at 21:26
  • $\begingroup$ and what about ^n $\endgroup$ 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: $$z^n = |z|^n(\cos(n\alpha)+i\sin(n\alpha))$$

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

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