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Possible Duplicate:
How do I get the square root of a complex number?

I'm currently playing with complex numbers and I realized that I don't understand how to compute $\sqrt{i + 1}$. My math program, sage, returns

$1.09868411346781 + 0.455089860562227*i$

How does sage compute that number? I don't even see how one could rewrite $\sqrt{i + 1}$ in a number of the form $a+bi$.

5 Answers 5

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Using the fact that $ {\rm e}^{i2k\pi} = 1 $ for any integer $k$, we have

$z^2 = {(1+i)} = \sqrt{2}\, { \rm e}^{i \frac{\pi}{4}} = \sqrt{2}\,{ \rm e}^{i \frac{\pi}{4}} { \rm e}^{2 k\pi i} = \sqrt{2}\,{ \rm e}^{i \left(\frac{\pi}{4} + 2k\pi\right)} $

$ \Rightarrow z = 2^{\frac{1}{4}} {\rm e}^{ \left(\frac{i\pi}{8} + k \pi i\right)} \,.$

Now, you can see that, if you take $k=0$, you will get the first root, and when you take $k=1$, you will have the second root. If you take $k \geq 2$, you will have the same first two roots. Then the only choices for $k$ is $0,1$.

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    @BabakS.: Thank you very much for the comment. I really appreciate it.2013-07-08
26

For finding $\sqrt{x+yi}$ put $\sqrt{x+yi}=a+bi$ 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 $(3):(a^2+b^2)^2=x^2+y^2$ Adding (1) and (3) gives you $2a^2=\sqrt{x^2+y^2}+x$ and subtracting (1) of (3) gives you $2b^2=\sqrt{x^2+y^2}-x$. Now, I think you can do your question by yourself. :)

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    Sir,i see that $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=x^2+y^2$.So $(3)$ equ. should be edited.2013-02-04
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$i + 1 = \sqrt2 \left ( {1 \over \sqrt 2} + {1 \over \sqrt 2} i\right ) \\ = \sqrt 2 \left( \cos \left( \pi \over 4 \right) + i \sin \left( \pi \over 4 \right) \right ) \\ = \sqrt 2 e^{i \pi \over 4}$

$\sqrt{i +1} =\left( 2\right)^{1 \over 4}e^{i \pi \over 8} = \left( 2\right)^{1 \over 4}\left( \cos \left( \pi \over 8 \right) + i \sin \left( \pi \over 8 \right)\right)$

Well, this is how Wolframalpha calculates. The other root would be $\left( 2\right)^{1 \over 4}\left( \cos \left( 9\pi \over 8 \right) + i \sin \left(9 \pi \over 8 \right)\right)$

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    Yep , thanks for that!!2012-08-28
10

Hint $\ $ Applying my Simple Denesting Rule and rationalizing denominators yields

$\rm \sqrt{1 + {\it i}\,}\, =\, \sqrt{\alpha}+ {\it i}\,\sqrt{-\alpha'}\quad for\quad \alpha,\alpha' \,=\, \frac{1\pm\sqrt{2}}2$

Indeed, squaring it yields $\rm\,\ \alpha+\alpha'\! + 2 \sqrt{-\alpha\alpha'}\:{\it i}\: =\, 1 + {\it i}$

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    @Serial Downvoter How about some constructive feedback?2012-08-28
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First, the notation $\sqrt{1+i}$ is misleading, because $i+1$ has two squared roots and there is no canonical way to distinguish one from the other. Indeed, $\displaystyle i+1= \sqrt{2} \left( \frac{ \sqrt{2}}{2}+ i\frac{\sqrt{2}}{2} \right)= \sqrt{2} e^{i \pi /4}$. So $\pm \ 2^{1/4} e^{i \pi /8}$ are the squared roots of $1+i$.

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    @celtschk What is not canonical is to decide that one wants a continuus square root on the complex plane except the negative real axis. One could also consider the complex plane except any ray issued from zero, or any other line issued from zero.2013-03-13