# Solving a quadratic equation with complex coefficient

Express $z4$=-$\sqrt{3}$+i in polar form. Hence solve the equation $Z^2$=$z4$ for $z$ a complex number. You may leave the answer in polar form.

My answer: $z4$ in polar form is 2cis-30$^{\circ}$ and thats as far as I have gotten. I have seen this question asked on other sites, but the answers with working solutions are too complex for the level this is at. This is a complex number topic assignment on a basic mathematics course, so I can only solve this question using the basic material that has been covered already in the course. There is an example in the notes where they use De Moivre's theorem to solve it.

• $$-\sqrt3+i=2\,\text{cis}\left(\arctan\frac1{-\sqrt3}\right)=2\,\text{cis}\,( 150^\circ)$$ and not $\;-30^\circ\;$ , since here cosine is negative and sine positive. – Timbuc Jun 30 '15 at 13:03
• $(a{\rm\ cis\ }b)^2=a^2{\rm\ cis\ }2b$ should give you an idea for how to take square roots of a number given in cis form. – Gerry Myerson Jun 30 '15 at 13:12

We have $z_4=2\mbox{cis}(150^\circ+360^\circ\times k), \ k=...-2,-1,0,1,2,...$, that is $\mbox{cis(.)}$ is periodic with period $360^\circ$. Then $z^2=z_4$ requires that $$z=\sqrt{2}\mbox{cis}\left(\frac{150^\circ+360^\circ \times k}{2}\right)\ k=...-2,-1,0,1,2,...$$ but there are only two distinct values for $z$ which correspond to $k=0,1$. Hence the two solutions are:$z=\sqrt{2}\mbox{cis}(75^\circ)$ and $z=\sqrt{2}\mbox{cis}(225^\circ)$