are these complex roots correct?

Question

I have a doubt about this exercise, it's about finding complex roots of a a complex equation (raised to a power greater than 1). I don't know if this method is correct, even if the solutions seem right.

I need to solve this exercise $$z^2 = -6$$

my attempt:

I used an external variable $w$, $w = z^2$ $$ w = -6 $$
now, I found a, b, rho, and theta. (rho is the norm of the complex number, and theta is the angle between the hypotenuse and the x-axis). $$a = -6\space (a\space is\space the\space real\space part)$$ and $$b = 0 \space(the\space imaginary\space part) $$ and now rho, and theta $$rho = ||-6|| = \sqrt(-6^2) = 6$$ and $$theta = arctan(b/a) = 0$$

okay, now I rewrote it in trigonometric form, and raising to 1/2 (because 2 is the power). $$w = \sqrt(6)*(cos(1/2 * 0+2k\pi)+sin(1/2 * 0+2k\pi)$$ and now I found easily complex roots, with k = 0, and with k = 1. $z = \sqrt(6) \space \space \space \space z = -\sqrt(6)$.

It cannot be correct because the solution that you found satisfy $z^2 =6$, and not $z^2=-6$. — Martin R, Jan 21 '22 at 08:41
You should get into the habit of checking if your solution really works. — Kavi Rama Murthy, Jan 21 '22 at 08:42
A good example that computing the argument of a complex number $z=x+iy$ via $\arctan(y/x)$ does not work in general. — Martin R, Jan 21 '22 at 08:43
@MartinR could you explain it better? because this is the method I learned during calculus 1 lectures, and I thought it always works — Gabriel Burzacchini, Jan 21 '22 at 08:46
My favourite language has the atan2 function which for atan2(0,-6) gives about $3.141593$ (your favourite might reverse these) — Henry, Jan 21 '22 at 08:46
@GabrielBurzacchini: $\arctan$ is always in the range $(-\pi/2, \pi/2)$, so that cannot work for complex numbers in the left half-plane or on the imaginary axis. — Martin R, Jan 21 '22 at 08:47

José Carlos Santos · Accepted Answer · 2022-01-21T08:51:09.877

1

That is wrong, since the square of each of those two numbers is $6$, rather than $-6$.

You have $-6=6(\cos(\pi)+\sin(\pi)i)$, and therefore a square root of $-6$ is$$\sqrt6\left(\cos\left(\frac \pi2\right)+\sin\left(\frac\pi2\right)i\right)=\sqrt6i,$$and another square root is$$\sqrt6\left(\cos\left(\frac{3\pi}2\right)+\sin\left(\frac{3\pi}2\right)i\right)=-\sqrt6i.$$

Of course, you can also say that, since the square roots of $-1$ are $\pm i$, and since $-6=\sqrt6^2\times(-1)$, the square roots of $-$ are $\pm\sqrt6i$.

edited Jan 21 '22 at 08:51

answered Jan 21 '22 at 08:44

José Carlos Santos

427,504

why you have 3/2? – Gabriel Burzacchini Jan 21 '22 at 08:50
Since $\cos(\pi)+\sin(\pi)i=\cos(3\pi)+\sin(3\pi)i$, a square root of $\cos(\pi)+\sin(\pi)i$ is$$\cos\left(\frac{3\pi}2\right)+\sin\left(\frac{3\pi}2\right)i.$$ – José Carlos Santos Jan 21 '22 at 08:52

are these complex roots correct?

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