1. ## Complex Number (Modulus/Phase)

I've been working with the equation below for too long that I'm forgetting the simplest operation. So...

$\frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2}$

Can the equation above by split into:

$\frac{z\theta_0^2}{\theta_0^2} - \frac{z\theta_0^2}{\theta^2}+ \frac{z\theta_0^2}{2i\theta\theta_0\phi}$?

2. Originally Posted by Air
$\frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2}$
Can the equation above by split into:
$\frac{z\theta_0^2}{\theta_0^2} - \frac{z\theta_0^2}{\theta^2}+ \frac{z\theta_0^2}{2i\theta\theta_0\phi}$?
NO!
It is that same as: $\frac{a}{b+c} \not= \frac{a}{b} + \frac{a}{c}$

3. Can I have hints on how to find the modulus?

I've been told that $\frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2}$ is a real number.

4. Can we write this as $\frac{{zA^2 }}{{ - B + 2iBAC + A^2 }}$
where A,B,C are real numbers?

5. Originally Posted by Plato
Can we write this as $\frac{{zA^2 }}{{ - B + 2iBAC + A^2 }}$
where A,B,C are real numbers?
Multiplying the numerator and the denominator by the conjugate of the denominator can get real and imaginary components. Then we can:

$\sqrt{(R)^2+(I)^2}$ where R is real number and I is the imaginary number co-efficient.

Would this work?

6. This is true.
$\frac{{zA^2 }}{{ - B + 2iBAC + A^2 }} = \frac{{zA^2 \left[ { - B - 2iBAC + A^2 } \right]}}{{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 }}$

7. Originally Posted by Plato
This is true.
$\frac{{zA^2 }}{{ - B + 2iBAC + A^2 }} = \frac{{zA^2 \left[ { - B - 2iBAC + A^2 } \right]}}{{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 }}$
And so...

$\frac{zA^2 \left[ { - B + A^2 } \right]}{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 } - \frac{zA^2 \left[ 2BAC\right]}{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 } i$