Question

Show that the complex nth roots of a nonzero complex number w are equally spaced on the circle.

Short answer

Arguments of the all roots are$\left(\frac{{\theta }_o}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{2\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{4\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{6\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{8\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{10\mathrm{\pi }}{m}\right)$Hence, the complex nth roots of a nonzero complex number w are equally spaced on the circle.

Step-by-step solution

Step 1. Considering a complex number w  

Assume $w=r(\cos {\theta }_{\circ }+i\sin {\theta }_{\circ })$
be a non zero complex number. According to the theorem it has m distinct roots where

$z_n=\sqrt[m]{r}\left[\left(\cos \left(\frac{{\theta }_o}{m}+\frac{2n\mathrm{\pi }}{m}\right)+i\sin \left(\frac{{\theta }_o}{m}+\frac{2n\mathrm{\pi }}{m}\right)\right)\right]$

n=0,1,2,3........(m-1)

Step 2. Conclusion

We can see that the argument of the complex number w is $\left(\frac{{\theta }_o}{m}+\frac{2n\mathrm{\pi }}{m}\right)$and this argument varies with value of n. All the arguments are

$\left(\frac{{\theta }_o}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{2\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{4\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{6\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{8\mathrm{\pi }}{m}\right),\left(\frac{{\theta }_o}{m}+\frac{10\mathrm{\pi }}{m}\right)$

So it is clear that all the arguments of the w are equally placed. When these roots plotted on the graph form a circle with origin as the centre and $\sqrt[m]{r}$is the radius .

Therefore the complex nth roots of a nonzero complex number w are equally spaced on the circle.