1. ## Complex Number Question

Hello there!

It's my first time here, so I apologize in not being able to find the 'Complex Numbers' part of these forums.

I must have forgotten how to solve these ones (ofcourse ), so here it is.

(x+ai)^2=3-4i - Solve for {a,x} given that they are both Real numbers.

Thanks, and if anyone knows the proper Complex Number forums page, please let me know

2. Originally Posted by Tan90
Hello there!

It's my first time here, so I apologize in not being able to find the 'Complex Numbers' part of these forums.

I must have forgotten how to solve these ones (ofcourse ), so here it is.

(x+ai)^2=3-4i - Solve for {a,x} given that they are both Real numbers.

Thanks, and if anyone knows the proper Complex Number forums page, please let me know
Either the Algebra or Pre-Calculus sub-forums are fine.

$(x + ai)^2 = 3 - 4i$

$x^2 + 2axi + a^2i^2 = 3 - 4i$

$x^2 + 2axi - a^2 = 3 - 4i$

$x^2 - a^2 + 2axi = 3-4i$.

Equating real and imaginary parts gives:

$x^2 - a^2 = 3$ and $2ax = -4$.

From the second equation we find $a = -\frac{2}{x}$.

Substituting into the first gives

$x^2 - \left(-\frac{2}{x}\right)^2 = 3$

$x^2 - \frac{4}{x^2} = 3$

$x^4 - 4 = 3x^2$

$x^4 - 3x^2 - 4 = 0$

$(x^2 - 4)(x^2 + 1) = 0$

$(x - 2)(x + 2)(x^2 + 1) = 0$.

Since we know $x$ is real, that means we can only accept the cases

$x - 2 = 0$ and $x + 2 = 0$.

Therefore $x = 2$ or $x = -2$.

Since we know $a = -\frac{2}{x}$

that means $a = -1$ or $a = 1$.

So the solutions are:

$(x, a) = (2, -1)$ and $(x, a) = (-2, 1)$.