# taking the absolute value of complex numbers to an arbitrary power [closed]

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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• A good starting point is to begin with $$i^1,i^2,i^3,i^4,i^5,...$$ – Dr. Sonnhard Graubner Jan 24 at 18:09
• Isn't $n$ a natural number? If so, then it for sure positive.. So $|n|=n$ – Fareed AF Jan 24 at 18:10
• 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. – José Carlos Santos Jan 24 at 18:10
• yes that is correct ^ – Jlatmer Jan 24 at 18:11
• If $z=x+iy$ is a complex number, $|z|=\sqrt{x^2+y^2}$. Now what are $x,y$ for $z=i$? – 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.
• How is it that you have gone from $|\frac{1}{n}|$ to $\frac{1}{|n|}$? Is this a rule? – Jlatmer Jan 24 at 18:23
• @Jlatmer you have that $| \frac{1}\{n}| = \frac{|1|}{|n|}$ (this is a rule), and $|1|=1$ – Alexandros Jan 24 at 18:25