Complex Numbers - Confused about how to go about question

Question

I would like some clarification about complex numbers. I have $p^3 = \frac{\sqrt{3} + i}{2}$ where $i = \sqrt{-1}$. If I was to find the modulus $r$ and arguments of the solution from $0≤θ<2π$ in polar form $re^{i\theta}$, would I have to firstly make $p$ the subject instead of $p^3$? Thanks

Answer 1

Not necessarily. Express $(\sqrt 3 + i)/2$ in the polar form first, $|r|e^{i \theta}$. Then

$$p^3 = \frac{\sqrt 3 + i}{2} = |r|e^{i \theta}$$

From there, finding $p$ is simple: take the cube root throughout and use known properties to simplify:

$$p = |r|^{1/3} \cdot e^{i \theta / 3}$$

If desired, convert your answer back into the Cartesian $x+iy$ form after this.

It is worth noting that three potential answers can result, since we're taking the cube root of $|r|$. In the complex plane, each root of $|r|$ lies on the circle of radius $|r|$, and are equidistributed along the circle (in the sense that they're all separated by angles of $120^\circ$); one will be a real number, and the other two will be complex. With this knowledge, though, it will at least be easy to ascertain which three values of $|r|^{1/3}$ you have, and in turn figure out the $p$ which satisfy your equation.

Answer 2

Claim: this answer concernes finding roots with DeMoivre's theorem.

The modulus of $z$ is $$|z|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=1$$ By DeMoivre's rule, we have: $$z=\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)=r^3(\cos(3\theta)+i\sin(3\theta))$$ The radius $r$ is $r=1$, while for $\theta$, we have: $$3\theta=\frac{\pi}{6}+2k\pi, k=0,1,2\leftrightarrow \: \theta_0=\frac{\pi}{18} \vee \theta_1=\frac{13\pi}{18} \vee \theta_2=\frac{25\pi}{18}$$

The three complex numbers are: $$p_0=\cos\left(\frac{\pi}{18}\right)+i\sin\left(\frac{\pi}{18}\right) \vee p_1=\cos\left(\frac{13\pi}{18}\right)+i\sin\left(\frac{13\pi}{18}\right) \vee p_2=\cos\left(\frac{25\pi}{18}\right)+i\sin\left(\frac{25\pi}{18}\right)$$