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# ​ x^3=-4+4i to polar form

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the question is to convert $${x^3}=-4+4i$$ into polar form.

I got this far:

$$x = \sqrt[3]{-4+4i}$$, so to convert the complex number inside the cubic root to polar coordinates,

$$r = \sqrt[2]{16+16}, \theta = arctan(-1)$$, which is $$r = \sqrt{32}, \theta =3pi /4$$,

since $$[rCis(\theta)]^n=r^n(cos(n\theta)+isin(n\theta)), n=1/3, r=\sqrt{32}$$,

$${\sqrt[2]{32}}^{1/3}(cis(3\pi/4*1/3))=\sqrt[6]{32}(cis(\pi/4))$$,

however to my surprise there are two more answers and I have no idea where else the answer can come from.

can someone find the error or sugget another way of getting it?

Jul 2, 2022
edited by Guest  Jul 2, 2022

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In the realm of complex numbers, every number has two square roots, three cube roots, four fourth roots, five fifth roots and so on.

Do what you've done except put the angle in its multivalued form,

$$\displaystyle 3\pi/4+2k\pi,$$ where k is any integer.

For cube roots, (after you've divided by three), let k = 0, 1, 2.

(Try k = 3 if you want but you should find that it just duplicates k = 0, (and k = 4 duplicates k = 1 etc.).)

If you were calculating square roots you would (after the division by 2) let k = 0 or 1, if calculating fourth roots, k = 0,1,2,3 and so on.

Jul 3, 2022