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. $z = \sqrt(6) \space \space \space \space z = -\sqrt(6)$.
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 $\pm\sqrt6i$.