Find all values in polar form.

Question

Studying for my engineering mathematics exam I can't figure out this question:

find all the values of (-3125)^1/5 in polar form?

should there be 5 values of (-3125)^1/5 in polar form?

my understanding is: -3125+j0= 3125∠0

Answer 1

Since $z=-3125$ is complex number, we can write it as $z=-3125 + 0 \cdot i$, i.e. we know it's real part is $-3125$, and it's imaginary part is $0$. Now we can write the given number $z$ in polar form: $z = \rho (\cos \varphi + i \sin \varphi),$ where $\rho = \sqrt{Re^2(z) + Im^2(z)}=3125,$ and $\varphi = \pi$ (trivial to prove), i.e. we have $$z = 3125(\cos \pi + i \sin \pi).$$

Now, we can apply well known formula for $n^{\text{th}}$ root of $z$, so we will have: $$z^{\frac{1}{5}} = \rho^{\frac{1}{5}}(\cos \frac{\varphi + 2k\pi}{n} + i \sin \frac{\varphi + 2k\pi}{n}) = 3125^{\frac{1}{5}}(\cos \frac{\pi + 2k\pi}{5} + i \sin \frac{\pi + 2k\pi}{5}), k=0,1,2,3,4.$$

There are 5 solutions: $$w_0=3125^{\frac{1}{5}}(\cos \frac{\pi}{5} + i \sin \frac{\pi}{5}) \\ w_1 = 3125^{\frac{1}{5}}(\cos \frac{3\pi}{5} + i \sin \frac{3\pi}{5}) \\ w_2 = 3125^{\frac{1}{5}}(\cos \pi + i \sin \pi) \\ w_3 = 3125^{\frac{1}{5}}(\cos \frac{7\pi}{5} + i \sin \frac{7\pi}{5}) \\ w_4 = 3125^{\frac{1}{5}}(\cos \frac{9\pi}{5} + i \sin \frac{9\pi}{5}). \\$$

$$w_0=5(\cos \frac{\pi}{5} + i \sin \frac{\pi}{5}) \\ w_1 = 5(\cos \frac{3\pi}{5} + i \sin \frac{3\pi}{5}) \\ w_2 = 5(\cos \pi + i \sin \pi) \\ w_3 = 5(\cos \frac{7\pi}{5} + i \sin \frac{7\pi}{5}) \\ w_4 = 5(\cos \frac{9\pi}{5} + i \sin \frac{9\pi}{5}).$$

Answer 2

Using that $-3125 = -5^5$, we can determine that the radii of our polar points are all 5.

Then, by the property of our roots all being equally spaced we can see that there must be an angle of $72^\circ$ between each root.

Thus the polar coordinates are $(5,180^\circ), (5, 108^\circ), (5, 36^\circ), (5, -36^\circ), (5, -108^\circ)$

Convert to radians if required.