Solve $w^2=−15+8i$, where $w$ is complex.
Normally, I would convert this into polar coordinates, but the problem is that is too slow.
What is another alternative?
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Caleb Stanford
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Amad27
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1Please see Bill Dubuque's easy method here: http://math.stackexchange.com/questions/646547/find-all-roots-of-z2-3-4i – Dec 11 '16 at 10:38
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How is converting to polar to slow? – bigfocalchord Dec 11 '16 at 11:17
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How is it fast? – Rainb Apr 25 '21 at 13:08
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Start out by letting $w = a + bi \implies w^2 = a^2-b^2 + 2abi = -15+8i \implies a^2-b^2 = -15, 2ab = 8 \implies a^2 - \dfrac{16}{a^2} = -15$. Put $x = a^2 \implies x - \dfrac{16}{x} + 15 = 0\implies x^2 + 15x - 16 = 0\implies (x-1)(x+16) = 0\implies x = 1\implies a = \pm 1, b = \pm 4$
DeepSea
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A general method that will very often work is to write all variables as the sum of the real and imaginary parts, and then equate the real and imaginary parts of the equation.
In this case, we set $w = a + bi$, and we get $$ (a + bi)^2 = -15 + 8i \\ $$ Then setting real and imaginary parts equal, \begin{align*} a^2 - b^2 &= -15 \tag{1} \\ 2ab &= 8 \tag{2} \end{align*} From (2), $b = \frac{4}{a}$; plug into (1) and multiply by $a^2$ we get $$ a^4 - 16 = -15a^2 $$ so $$ (a^2 + 16)(a^2 - 1) = 0 $$ and $a, b \in \mathbb{R}$. So $a = 1$ or $a = -1$. And that will make $b = 4$ or $b = -4$, respectively.
Caleb Stanford
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