Solving equation with complex numbers My lecturer presented a equation with complex numbers that he simplified by completing the square to the following:
$$(z + (i-1))^2 = -3+4i$$
Next he set $$w = z + (i-1) \\ w^2 = -3+4i$$
My first question is why he did that? Why does it make the equation easier to solve? If would just be this if he did not introduce $w$:
$$z = \pm \sqrt{-3+4i} + (i - 1)$$
After introducing $w$ he says $$w = 1+2i \ \text{or} \ w = -1 - 2i$$
I'm not sure if my notes are missing something, but how did he solve $$w^2 = -3+4i$$
How does he take the square root of $-3+4i$?
 A: $w$ is nothing but an auxiliary unknown $\sqrt{-3 + 4i}$, but it makes the argument more clear by explicitly find the standard complex form of $\sqrt{-3 + 4i}$. In fact, let
$$ w = u + iv, u, v \in \mathbb{R} $$
Then we have 
$$ -3 + 4i = w^2 = (u^2 - v^2) + 2iuv $$
On solving
$$ u^2 - v^2 = -3 $$
$$ 2uv = 4 $$
Multiplying the first equation by $u^2$ and substituting the second into the first. Then it becomes
$$ u^4 - 4 = -3u^2$$
which is a quadratic equation in $u^2$. Eventually we obtain
$$ u^2 = 1 \quad \textrm{or} \quad u^2 = -4 \quad \textrm{(rejected)}$$
$$ u = 1, v = 2 \quad \textrm{or} \quad u = -1, v = -2$$
i.e. $w = 1+2i \quad \textrm{or} \quad w = -1 - 2i$
A: Hint
Let us consider the equation $$w^2=-3+4i$$ and write $w=a+i b$; expanding, we then have $$a^2-b^2+2abi=-3+4i$$ which implies $$a^2-b^2=-3$$ $$2ab=4$$ Use the second equation to eliminate $b$ using then $b=\frac 2a$ we plug in the first to get $$a^2-\frac{4}{a^2}=-3$$ Multiply everything by $a^2 $ and get $$a^4+3a^2-4=0$$ which is a quadratic equation in $a^2$.
I am sure that you can take from here.
A: Let $a+bi$ be one of squere root of $-3+4i$. Then:
$$(a+bi)^2=a^2-b^2+2abi$$
So:
$$a^2-b^2=-3$$
$$2ab=4$$
In this case it's easy to guess such $a,b$: $a=1, b=2$ or $a=-1,b=-2$ (so I think that the lecturer guessed the solutions).
Introduction $w=z+(1+i)$ only simplify the notation.
