A complex number $u$ is given by $u = -1+(4\sqrt{3})i$. Find the two square roots of $u$.
Now, I know we have to compare the equation with $a+bi$ but my text book doesn't square both sides in the equation $-1+(4\sqrt{3})i = a+bi$. Only the latter is squared.
Can you explain why?
On
To solve $(a+bi)^2=c+di$ use $a^2-b^2=c,\,2ab=d$ to deduce $a^2+b^2=\sqrt{c^2+d^2}$ so $a^2=\frac{\sqrt{c^2+d^2}+c}{2},\,b^2=\frac{\sqrt{c^2+d^2}-c}{2}$. Of the four choices of $a,\,b$ satisfying these expressions for $a^2,\,b^2$, only two are consistent with $ab=\frac{d}{2}$. In this case $c=-1,\,d=4\sqrt{3}$ so $\sqrt{c^2+d^2}=7$ and $a^2=3,\,b^2=4$, and since $ab=2\sqrt{3}$ one square root is $\sqrt{3}+2i$.
The fast way uses the remark that, if $z=a+ib$, then $|z|=\sqrt{|u|}$, or $|z|^2=|u|$. So you have the equations \begin{align} &a^2-b^2=-1 \tag{1}\\ &2ab =4\sqrt 3 \tag{2}\\ &a^2+b^2=\sqrt{(-1)^2+16\cdot 3}=7\tag{3} \end{align} Equations $(1)$ and $(3)$ yield instantly $$\begin{cases} 2a^2=6\\ 2b^2=8 \end{cases}\iff a=\pm\sqrt {3},\quad b=\pm2.$$ Using eq. $(2)$, you know $a$ and $b$ have the same sign. Hence $$z=\pm(\sqrt {3}+2i).$$