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I was reading this question earlier: Understanding imaginary exponents

In the answer, the answerer says

$A^i=x+iy$

Furthermore, we can write $A^{−i}=x−iy$ for the same $x$ and $y$.

Can someone explain to me why this is true? It kind of makes sense to me but how can we make the leap to just negating the imaginary component?

Community
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Inbl
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2 Answers2

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If $A>0$ is real, we define $A^z=e^{z\ln A}$. So

$A^i=e^{i\ln A}=\cos(\ln A)+i\sin(\ln A)$ $A^{-i}=e^{-i\ln A}=\cos(-\ln A)+i\sin(-\ln A)=\cos(-\ln A)-i\sin(\ln A)$

John
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Hint: $$A^i=x+iy$$ $$A^{-i}=\frac{1}{A^{i}}=\frac{1}{x+iy}=\frac{1}{x+iy}\cdot\frac{x-iy}{x-iy}=\frac{x-iy}{x^2+y^2}=x-iy$$

Note : it says $$|A^i|=1$$

$$|A^i|=\sqrt{x^2+y^2}=1$$ $$\Rightarrow x^2+y^2=1$$

  • Thanks! Do you know why the note is true? Also, it seems like the answerer in the question I referenced was using this to prove that x^2+y^2=1, isn't this a circular proof? – Inbl Oct 24 '14 at 07:49
  • @Inbl I think That is clearly mentioned in that answer –  Oct 24 '14 at 08:03
  • So because |A^i| = 1, we can show x^2+y^2 = 1. We then use that to show that A^-i = x-iy? – Inbl Oct 24 '14 at 16:00