Find $x$ such that $\frac{(1+ 2i)^2}{x+i}$ is purely imaginary

Question

If $x$ is a real number, $\displaystyle\frac{(1+ 2i)^2}{x+i}$ is purely imaginary, what is the value of $x$?

So I expanded the numerator to $-3+4i$, which turns the imaginary number into $\displaystyle\frac{-3+4i}{x+i}$ from here though I'm not sure where to go to figure out what $x$ is.

Thanks for any direction or help.

Answer 1

$$\frac{-3+4i}{x+i}=ki \implies -3+4i = kxi -k \implies k=3 \text{ and } x=\frac 4 3$$

Answer 2

Geometrically, you want $x+i$ to be at right angles to $-3+4i$.

Draw a diagram and find the intersection between the line with imaginary part $1$ and the ray from $0$ in the direction $4+3i$.

You get $x=4/3$.

Answer 3

Multiply top and bottom by the conjugate of the denominator so that denominator becomes purely real.

Answer 4

Hint multiply by $\frac{x-i}{x-i}$

Answer 5

As $a+ib+(a-ib)=2a=0$ if $a+ib$ is purely imaginary

$$\implies\dfrac{-3+4i}{x+i}+\dfrac{-3-4i}{x-i}=0$$

$$\iff(-3+4i)(x-i)+(-3-4i)(x+i)=0$$

$$\iff-6x-4i(i)=0\iff x=?$$