I need $|\frac{i^{n}}{n}|$ and I have seen the problem simplified to $\frac{|i^{n}|}{n}$ and I am confused by this as isn't $\frac{1}{n}$ the coefficient of i so we could just square it and take the square root to find the absolute value. Further I do then not understand how $|i^{n}| = 1$


closed as off-topic by José Carlos Santos, Namaste, B. Goddard, rtybase, Alexander Gruber Jan 24 at 23:33

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  • $\begingroup$ A good starting point is to begin with $$i^1,i^2,i^3,i^4,i^5,...$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 24 at 18:09
  • $\begingroup$ Isn't $n$ a natural number? If so, then it for sure positive.. So $|n|=n$ $\endgroup$ – Fareed AF Jan 24 at 18:10
  • 1
    $\begingroup$ Is the equality $\left\lvert\frac{i^n}n\right\rvert=\frac{\lvert i^n\rvert}n$ that confused you? Your question is not clear. $\endgroup$ – José Carlos Santos Jan 24 at 18:10
  • $\begingroup$ yes that is correct ^ $\endgroup$ – Jlatmer Jan 24 at 18:11
  • $\begingroup$ If $z=x+iy$ is a complex number, $|z|=\sqrt{x^2+y^2}$. Now what are $x,y$ for $z=i$? $\endgroup$ – James Jan 24 at 18:27

I am not quite sure I get your question. What do you mean by "isn't $\frac{1}{n}$ the reciprocal of i? In general, it holds that: $\vert z_1z_2 \vert = \vert z_1\vert\vert z_2\vert$ and $\vert z_1^{n} \vert = \vert z_1 \vert^{n}$. If $z_1 = \frac{1}{n}, z_2=i^{n}$ and use of property 1 followed by property 2 gives: $ \frac{|i|^{n}}{|n|}$, and I take it that n is non-negative so you might remove the modulus from the denominator.

  • $\begingroup$ Apologies I meant coefficient. $\endgroup$ – Jlatmer Jan 24 at 18:17
  • $\begingroup$ The absolute value of any power of i is 1. $\endgroup$ – user247327 Jan 24 at 18:20
  • $\begingroup$ How is it that you have gone from $|\frac{1}{n}|$ to $\frac{1}{|n|}$? Is this a rule? $\endgroup$ – Jlatmer Jan 24 at 18:23
  • $\begingroup$ @Jlatmer you have that $| \frac{1}\{n}| = \frac{|1|}{|n|}$ (this is a rule), and $|1|=1$ $\endgroup$ – Alexandros Jan 24 at 18:25

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