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The above expression popped up in a problem I was solving. Using the relation $$i=\exp(i\pi/2)$$ I got $$\frac{1}{\sqrt{\pm i}}=\exp(\mp i\pi/4),$$ however, the reference I have gives the result as $$\frac{1}{\sqrt{\pm i}}=\exp(\pm i\pi/4).$$ Please could someone point out my mistake or show how to properly evaluate the expression.

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$\sqrt {\pm i}$ has four values, so you're both wrong. –  Chris Eagle Jan 4 '13 at 13:47
    
@ChrisEagle: Okay. Can you please show how to evaluate? –  Jonjo Jan 4 '13 at 13:53
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3 Answers

up vote 5 down vote accepted

$i$ has modulus $1$ and argument $\pi/2$, so one of the square roots of $i$ has modulus $1$ and argument $\pi/4$, while the other is the negative of this one, so it has modulus $1$ and argument $5\pi/4$. Then the two values of $1/\sqrt i$ have modulus $1$ and arguments $-\pi/4$ and $-5\pi/4$.

Similarly approach for $-i$, which has modulus $1$ and argument $3\pi/2$.

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Mine is sooooo elementary than yours, Gerry and Clive. :( –  B. S. Jan 4 '13 at 13:59
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@Babak, whatever floats your boat. I'm not sure what's non-elementary about modulus and argument. –  Gerry Myerson Jan 4 '13 at 14:08
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I am trying to show you how to find $z=\sqrt{x+yi}$. Let $z=\sqrt{x+yi}=a+bi$ so $$(\sqrt{x+yi})^2=(a+bi)^2$$ If you do the latter identity, you will find $$(1): a^2-b^2=x;(2):2ab=y$$ and $$(3):(a^2+b^2)^2=\sqrt{x^2+y^2}$$ $(1)+(3)$ gives you $$2a^2=\sqrt{x^2+y^2}+x$$ and $(1)-(3)$ gives you $$2b^2=\sqrt{x^2+y^2}-x$$ Now, the only thing you should kow is what is $\frac{1}{a+ib}$

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$$\pm i=e^{\pm\pi/2}\Longrightarrow \sqrt{\pm i}=\pm e^{\pm\,\pi/4}\Longrightarrow\frac{1}{\sqrt{\pm i}}=\left(\sqrt{\pm i}\right)^{-1}=$$

$$\pm e^{\mp \pi/4}=\pm\left(\cos\frac{\pi}{4}\mp i\sin\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt 2}\left(1\mp i\right)$$

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