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I want to find the square roots of a complex number, $w = a+ib \in \mathbb{C}$, i.e. I'm looking for solutions, $z = x + iy$, for the equation $z^2 = w$.

This question has been asked here a couple of times, but I still don't get why there are only the two solutions, $$ z = \pm \left( \sqrt{\frac{|w|+a}{2}} + i \text{sgn}(b)\sqrt{\frac{|w|-a}{2}}\right) $$.

How do I see that my intermediate results $$x=\pm \sqrt{\frac{|w|+a}{2}}\\ y=\pm \sqrt{\frac{|w|-a}{2}}$$ cannot be combined to yield four solutions? I feel like I am missing something quite elementary here.

Peter Mortensen
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Marc
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    Try squaring them, and see that only 2 of the 4 combinations work. – Gerry Myerson Apr 27 '15 at 12:51
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    Very often the "solution procedure" of an algebraic problem introduces new "roots" that don't actually satisfy the original equation. These are called "artifact roots" because they were introduced by the solution procedure, but one always is obliged to check (as Gerry advises) to see which of them satisfy the original equation. – hardmath Apr 27 '15 at 13:01
  • I'll keep this in mind! – Marc Apr 27 '15 at 13:03
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    To put it more precisely, it is simply an issue of logic. You have shown something of the form "If $(x+yi)^2 = a+bi$, then ...". Of course the converse may not hold. – user21820 Apr 28 '15 at 06:37
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    you can also directly implement the formula "square root of radius times half the angle" as $$\pm\sqrt{|z|}\frac{|z|+z}{|,|z|+z,|}.$$ Numerical stability is only guaranteed for $Re(z)>0$, use $\sqrt{z}=i\sqrt{-z}$ for the other half. – Lutz Lehmann Apr 28 '15 at 09:25

3 Answers3

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The identification $(x+\mathrm i\mkern1mu y)^2=a+\mathrm i\mkern1mu b$ leads to the equality $2xy=b$, hence the signs of $x$ and $y$ are the same if $b> 0$, opposite if $b<0$. This shows you cannot combine the signs of the real and imaginary parts of $z$ in an arbitrary way.

Bernard
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In a field, such as the complex numbers, there can be at most two square roots of any element. This is a consequence of the Fundamental Thm. of Algebra, which tells us that an $n$-th degree polynomial has at most $n$ roots.

Here the application is to a quadratic polynomial $p(x) = x^2 - a$ for field element $a$. Any square root of $a$ is a "root" (zero place) of the polynomial $p(x)$.

So in particular complex numbers have at most two square roots. As it happens, the only complex number which does not have two different square roots is zero, since any square root has an additive inverse that is also a square root.

hardmath
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  • Thank you for answering! I know the fundamental theorem, I just didn't see how to reduce the number of solutions. – Marc Apr 27 '15 at 13:02
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Understanding squaring and taking roots in the complex numbers (or more general anything about multiplication that does not involve addition) is easier to understand in the 'polar coordinate' picture, i.e. using the length/modulus (distance to zero) and argument (angle between the pos real axis and the line from 0 to the number) of the number rather than the real and imaginary part.

If we multiply two complex numbers their arguments add up and their lengths multiply. Now we look for numbers $w$ such that $w^2 = z$. This means that the length of $w$ multiplied by itself is the length of $z$ and that the argument of $w$ added to itself (i.e. multiplied by 2) equals the argument of $z$.

It follows that $|w| = \sqrt{|z|}$ (here we can only use the positive root because it is a length) and $\arg(w) = 1/2 \arg(z)$, or - surprise twist - $\arg(w) = 1/2 \arg(z) + \pi$ (because $2\pi = 0$ in terms of angles). This is all there is.

(I am a aware that this is more an answer to the question in the title than to the one in the body of your post)

Vincent
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