How to compute $\sqrt{i + 1}$ [duplicate]

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 $a+bi$.

Solution 1:

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. :)

Solution 2:

$i + 1 = \sqrt2 \left ( {1 \over \sqrt 2} + {1 \over \sqrt 2} i\right ) \\ = \sqrt 2 \left( \cos \left( \pi \over 4 \right) + i \sin \left( \pi \over 4 \right) \right ) \\ = \sqrt 2 e^{i \pi \over 4}$

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

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

Solution 3:

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

Solution 4:

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 $1+i$.

Solution 5:

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 $0,1$.