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 $-3+4i$?
Answer
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.
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