im reading the book "what is mathematics" and find the questions.

Calculate $\sqrt{5+12i}$

i followed the hint and wrote the equation $\sqrt{5+12i} = x+yi$ and solved it with these results :

  1. x=3,y=2
  2. x=-3,y=-2

so here my intuition told me maybe I should pick the 3+2i.
but I really don't know why?How to know whenther the complex number is >0 or <0 ?
or maybe I was wrong?

  • $\begingroup$ This is actually something you're probably familiar with. To phrase differently, is $\sqrt{9}=3$, or perhaps $\sqrt{9}=-3$? The truth is that you could pick either, and indeed they are the two roots of the polynomial $x^2-9$. It is a (the) fundamental theorem of algebra that every polynomial of degree $2$ over the complex numbers has two complex roots, and you've found both for the polynomial $z^2 - (5+12i)$. $\endgroup$ – Jonathan Y. Oct 12 '13 at 11:54
  • $\begingroup$ Both answers are valid, just how $-1$ and $1$ are both square roots of $1$. $(-3 - 2i)^2 = (-1)^2(3 + 2i)^2 = (3 + 2i)^2 $ $\endgroup$ – MCT Oct 12 '13 at 13:02

The function $\sqrt{\ }$ is not defined on $\mathbb C$, roughly for the reason you explain. Note that it is not even defined on $\mathbb R$. What is well defined is a function $\sqrt{\ }:\mathbb R_+\to\mathbb R_+$, which everybody knows.

Thus, "compute $\sqrt{x}$" when $x$ is in $\mathbb C\setminus\mathbb R_+$ can only mean providing the two complex numbers $z$ and $-z$ such that $z^2=x$. If $x=5+12\mathrm i$, you showed that $\{z,-z\}=\{3+2\mathrm i,-3-2\mathrm i\}$.

Edit: As mentioned in the comments, one can define a function $v$, continuous on the angular sector $(-\pi,\pi]$, and such that $v(z)^2=z$ for every $z$. Each $z$ in $\mathbb C$, $z\ne0$, can be uniquely written as $z=r\mathrm e^{\mathrm it}$ for some $r\gt0$ and $-\pi\lt t\leqslant\pi$. Define $v(0)=0$ and $v(z)=\sqrt{r}\cdot\mathrm e^{\mathrm it/2}$ for every such $z\ne0$.

Note that $v$ is continuous but for a twisted topology of the complex plane $\mathbb C$, where the (small) neighborhoods of every $z$ not in $\mathbb R_-^*$ are the usual (small) ones and the (small) neighborhoods of $z$ in $\mathbb R_-^*$ are the intersections of the usual (small) ones with the halfplane $\Im\geqslant0$. In particular, for every $r\gt0$, $v(r\mathrm e^{\mathrm it})\to-\mathrm i\sqrt{r}$ when $t\to-\pi$, $t$ in $(-\pi,\pi]$, while $v(r\mathrm e^{\mathrm it})\to+\mathrm i\sqrt{r}$ when $t\to\pi$, $t$ in $(-\pi,\pi]$.

  • $\begingroup$ In complex analysis, we sometimes select a distinguished choice for $\sqrt{\ }$. Probably the most common is: the square-root with positive real part, or {zero real part and positive imaginary part}. $\endgroup$ – GEdgar Oct 12 '13 at 12:13
  • $\begingroup$ @GEdgar This choice corresponds to defining $\sqrt{\ }$ continuously on the angular sector $(-\pi,\pi]$. $\endgroup$ – Did Oct 12 '13 at 12:20
  • $\begingroup$ @Did hi,I'm really curious about how you choose the z and -z,could you explain a little about the defining √ continuously on the angular sector (−π,π] ? $\endgroup$ – Detective King Oct 12 '13 at 13:42
  • $\begingroup$ @user2166063 See Edit. $\endgroup$ – Did Oct 12 '13 at 13:59
  • $\begingroup$ i have to say this is beyond my knowledge.Maybe someday I looked back and I could figure it out.But still thanks for the explaination. $\endgroup$ – Detective King Oct 12 '13 at 14:34

“Calculate $\sqrt{5+12i}$” really just asks to find a complex number $z$ with $z^2=5+12i$, which you did. And you noticed that there is more than one such number, because if $z$ is a solution, then so is $-z$, since of course $(-z)^2=z^2$. Now you have learned that when calculating the square root of a positive real number then you should always pick the positive solution, but that is just a convention. Indeed it is a convention that mathematicians will not always follow, as long as that does lead to confusion. You ask what positive and negative mean for complex numbers, and indeed it does not mean anything, therefore this convention would not make any sense here.


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