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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 $a+bi$.

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    $\begingroup$ $1.09868411346781 = \sqrt{\frac{\sqrt{2}+1}2} $ and$ 0.455089860562227 = \sqrt{\frac{\sqrt{2}-1}2} $ $\endgroup$ Feb 28, 2019 at 21:45

5 Answers 5

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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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  • $\begingroup$ Remark that this is a special case of the Simple Denesting Rule linked in my answer. $\endgroup$ Aug 28, 2012 at 14:26
  • $\begingroup$ @BillDubuque: Thanks for noting me that. I was not aware of your clever answer here. Mine looks so elementary. :) $\endgroup$
    – Mikasa
    Aug 28, 2012 at 14:32
  • $\begingroup$ The Simple Denesting Rule may be viewed as a convenient way to remember the solution of the above system of equations. $\endgroup$ Aug 28, 2012 at 15:05
  • $\begingroup$ 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. $\endgroup$
    – user52976
    Feb 4, 2013 at 17:14
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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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  • $\begingroup$ But you might add that one could also write it as $\sqrt2e^{i(\pi/4+2\pi)}$, which yields the other square root, ${(2)}^{1/4}e^{9i\pi/8}$. $\endgroup$
    – MJD
    Aug 28, 2012 at 9:45
  • $\begingroup$ Yep , thanks for that!! $\endgroup$
    – S L
    Aug 28, 2012 at 10:18
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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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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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  • $\begingroup$ I think the canonical choice is the one with positive real part, unless the real part is $0$, then it's the one with positive imaginary part. Or more generally, if you write the number as $r\mathrm e^{\mathrm i\phi}$ with $-\pi < \phi \le \pi$, then the canonical choice is $\sqrt{r}\mathrm e^{\mathrm i\phi/2}$ (and $r^{1/n}\mathrm e^{\mathrm i\phi/n}$ for the $n$th root). Note that without a canonical choice, $\sqrt{2}$ would be just as ambiguous. $\endgroup$
    – celtschk
    Aug 28, 2012 at 12:40
  • $\begingroup$ It corresponds to the lexicographic order, but since $\mathbb{C}$ is not an ordered field, I am not sure that it is really canonical. $\endgroup$
    – Seirios
    Aug 28, 2012 at 13:48
  • $\begingroup$ That it corresponds to the lexicographic order is coincidential. What is not coincidental is that for positive reals it corresponds with the choice made in real numbers. Also a wanted property is that it is continuous except for the non-positive reals. And I guess it's also a wanted property that for all numbers except of the negative reals, the square root of the conjugate is the conjugate of the square root (for the negative reals, it's not possible to achieve that). $\endgroup$
    – celtschk
    Aug 28, 2012 at 14:51
  • $\begingroup$ Also note that the square root (as I wrote in the first comment) is just the special case of the more general $n$th root, where there's the special property that all $n$th roots of $1$ can be obtained as powers of the canonical one (which is not the case for all possible choices if $n$ is compound). $\endgroup$
    – celtschk
    Aug 28, 2012 at 14:56
  • $\begingroup$ @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. $\endgroup$
    – Did
    Mar 13, 2013 at 23:14
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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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  • $\begingroup$ @BabakS.: Thank you very much for the comment. I really appreciate it. $\endgroup$ Jul 8, 2013 at 18:36

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