Complex Number Question

**Tan90** · April 7th 2010, 09:09 PM

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

**Prove It** · April 7th 2010, 10:14 PM

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

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