Understand roots of $(z+i)^2=(\sqrt3+i)^3$

Question

I'm trying to solve the equation $$(z+i)^2=(\sqrt3+i)^3$$ but I don't know how to extract the roots $$(z+i)^2=(\sqrt3+i)^3 \rightarrow (z+i)^2=8i \rightarrow z^2+(2i)z-(8i+1)=0$$

$z_{1,2}=-i \pm \sqrt{8i}$.

According to my book the solutions are $2+i$ and $-2-3i$ and I think that they are another way to write them but I don't know how to have them. Can someone help me to understand?

Answer 1

Recast your results,

$$-i \pm \sqrt{8i}=-i \pm 2\sqrt{2i}=-i \pm 2\sqrt{(1+i)^2}=-i \pm 2(1+i)$$

which are $2+i$ and $-2-3i$.

Answer 2

We have that

$$(z+i)^2=(\sqrt3+i)^3=8i=8e^{i\left(\frac \pi 2+2k\pi\right)}\implies z+i=2\sqrt 2e^{i\left(\frac \pi 4+k\pi\right)}$$

for $k=0,1$ that is

• $z+i=2\sqrt 2e^{i\frac \pi 4}=2+2i \implies z=2+i$
• $z+i=2\sqrt 2e^{i\frac {5\pi} 4}=-2-2i \implies z=-2-3i$

Answer 3

If you note $(2+2i)^2=8i$, you'll see your answer matches the book's.

Answer 4

You have $z_{1,2} = -i + K$ where $K^2 = 8i$.

So find $K$ where $K^2 = 8i$.....

Let $K = a+bi$ so $K^2 = (a^2 -b^2) + 2abi = 0 + 8i$ so $a^2-b^2 =0$ and $2ab=8$

So solve for $a$ and $b$.....

$a^2 - b^2 =0$ means $a=\pm b$ and $2ab = 8$ means $ab=4$ means $a$ and $b$ are both the same sign. So $a=b$ and $ab =a^2=4$ so $a =\pm 2$ and $b=\pm 2$.

So $K = \pm (2 + 2i)$ and $z_{1,2}=-i \pm(2+2i) = \begin{cases}2+i\\-2-3i\end{cases}$.