Solving equation with complex numbers

Question

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 1

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.

Answer 2

$w$ 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. $w = 1+2i \quad \textrm{or} \quad w = -1 - 2i$

Answer 3

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.