# Complex numbers exercise

## 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?

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.

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.

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

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

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?

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: $$2ab=4$$

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?

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.

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!

eumyang
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?
$$a^4 - 3a^2 - 4 = (a^2 - 4)(a^2 + 1)$$