- #1

- 539

- 246

- TL;DR Summary
- If it comes to calculate the principal square root of a complex number, it yields a unique number. Is it because it calculates two squares at the same time, that is, because includes that of a +bi and -a-bi?

This is a quote from "Calculus", by Robert A. Adams. It's a translation from spanish:

"

If ##a## is a positive real number, there exist two different real numbers whose square is ##a##. They are

Be ##r=|z|##, so ##r>0##. Be ##\theta=\mbox{Arg}(z)##. Hence, ##-\pi<\theta\leq{\pi}##. As

It's remarkable that the real part of ##\sqrt{z}## is always non-negative because ##\cos{(\theta/2)}\geq{0}## in ##-\pi/2<\theta\leq{\pi/2}##. At this interval ##\sin{(\pi/2)}=0## only if ##\theta=0##, in which case ##\sqrt{z}## is real and positive."

The questions are: ##\sqrt{z}## is at the same time square root of ##w_1## and ##w_2=-w_1##?; and why do we choose only the positive square ##\sqrt{z}##?

Thanks! I'm not native, so forgive the language mistakes. Of course, if it doesn't make sense anything, ignore me. I've posted maybe too quickly.

"

**Roots of square numbers**If ##a## is a positive real number, there exist two different real numbers whose square is ##a##. They are

##\sqrt{a}\;## (the positive square root of ##a##)

##-\sqrt{a}\;## (the negative square root of ##a##)

Every complex number ##z=x+yi## different from zero (this is, ##x^2+y^2>0##) has two square roots; if ##w_1=z## is a complex number so that ##w_1^2=z## then ##w_2=-w_1## also suits ##w_2^2=z##. Is interesting to point out one of this roots to call it ##\sqrt{z}##.##-\sqrt{a}\;## (the negative square root of ##a##)

Be ##r=|z|##, so ##r>0##. Be ##\theta=\mbox{Arg}(z)##. Hence, ##-\pi<\theta\leq{\pi}##. As

##z=r(\cos{\theta}+i\sin{\theta})##

the complex number##w=\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)##

fills clearly ##w^2=z##. We will call ##w## the **principal square root**of ##z##, and we will write it as ##\sqrt{z}##. The two solutions of the equation ##w^2=z## are, thus, ##w=\sqrt{z}## and ##-\sqrt{z}##.It's remarkable that the real part of ##\sqrt{z}## is always non-negative because ##\cos{(\theta/2)}\geq{0}## in ##-\pi/2<\theta\leq{\pi/2}##. At this interval ##\sin{(\pi/2)}=0## only if ##\theta=0##, in which case ##\sqrt{z}## is real and positive."

The questions are: ##\sqrt{z}## is at the same time square root of ##w_1## and ##w_2=-w_1##?; and why do we choose only the positive square ##\sqrt{z}##?

Thanks! I'm not native, so forgive the language mistakes. Of course, if it doesn't make sense anything, ignore me. I've posted maybe too quickly.