# Thread: Another Complex Number Question

1. ## Another Complex Number Question

If $z=x+yi$ determine the values of $x$ and $y$ such that $z=\sqrt{3+4i}$

2. $3+4i=4+4i-1=4+4i+i^2=(2+i)^2$

Then $z_1=2+i, \ z_2=-2-i$

3. Cool. Thanks for your help.

4. Originally Posted by Stroodle
If $z=x+yi$ determine the values of $x$ and $y$ such that $z=\sqrt{3+4i}$
Another approach:

$z^2 = 3 + 4i$

$\Rightarrow (x + iy)^2 = 3 + 4i$

$\Rightarrow (x^2 - y^2) + 2xy i = 3 + 4i$.

Therefore:

$x^2 - y^2 = 3$ .... (1)

$2xy = 4 \Rightarrow xy = 2 \Rightarrow y = \frac{2}{x}$ .... (2)

Substitute (2) into (1): $x^2 - \frac{4}{x^2} = 3 \Rightarrow x^4 - 3x^2 - 4 = 0 \Rightarrow (x^2 - 4)(x^2 + 1) = 0$.

But since x is real, $x = \pm 2$.

x = 2: y = 1.
x = -2: y = -1.

5. Awesome. Thanks for that.
This method will be much easier for me to reproduce for similar problems.