# Complex numbers exercise

renvox

## Homework Statement

Hi,
I have to solve this exercise:
"Given that (a+bi)^2 = 3+4i obrain a pair of simultaneous equations involving a and b. Hende find the two square roots of 3+4i. Hence find the two suqre roots of 3+4i."
I don't really know to do do it.

2. The attempt at a solution
Well, that's what I thought:
(a+bi)^2 = 3+4
we expand:
a^2+2abi-b^2=3+4i
we group them:
2abi-4i=-a^2+b^2+3
common factor:
2i(ab-2)=-ab^2+b^2+3
2i=(-a^2+b^2+3)/(ab-2)

And now I don't even know if what I've done is good, and if it's good what should be the next step?

Bohrok
Starting at a2 + 2abi - b2 = 3 + 4i, you want to equate the real and imaginary parts.
For example, if c + di = x + yi, then c = x and d = y. This way you'll get a system of equations to solve.

Karlx
Hi renvox.

Your initial equation $$\left(a+bi\right)^{2}=3+4i$$ involves the equality of two complex numbers. This implies that both the real part and the imaginary part of the two numbers must be equal.

renvox
Ok, so it will be:
2abi=4i
and
a^2-b^2=3
And now what do I do? :s

Karlx
You have a system of two equations and two variables...

renvox
Sorry, it's my second day with complex numbers.
2abi=4i
a=sqrt(3+b^2)
therefore:
2bi*sqrt(3+b^2)=4i
-4bi*(3+b^2)=-16
-12b^2-4b^4=-16
dividing by -4
b^4+3b^2=4
common factor b^2
b^2(b^2+3)=4
right?

Karlx
The system of equations you must solve is the following:

From the equality of the real parts: $$a^{2}-b^{2}=3$$
From the equality of the imaginary parts: $$2ab=4$$

renvox
lol, i'm making new laws of complex numbers :x
so:
b=2/a
a^2-2/a=3
a^3-3a-2=0
do I have to use cubic equation or is there another way to do it?

Bohrok
Make sure you're doing your substitution correct. In a2 - b2 = 3, you have to square the 2/a when you substitute it in for b. So you'll get
$$a^2 - \left(\frac{2}{a}\right)^2 = 3$$

Try taking it from there.

renvox
thanks guys, you are awesome.
a^2-(4/a^2)=3
a^4 - 3^2 -4 = 0
by plugging: a=-2 or 2
so b= 1 or -1
thanks!

Homework Helper
thanks guys, you are awesome.
a^2-(4/a^2)=3
a^4 - 3^2 -4 = 0
Forgot an a2. This should be
$$a^4 - 3a^2 - 4 = 0$$
by plugging: a=-2 or 2
By guess-and-check? You realize this is factorable, right?
$$a^4 - 3a^2 - 4 = (a^2 - 4)(a^2 + 1)$$