# Vectors and imaginary numbers

1. Aug 27, 2013

### bfusco

1. The problem statement, all variables and given/known data
Consider a vector z defined by the equation z=z1z2, where z1=a+ib, z2=c+id.
(a) show that the length of z is the product of the lengths of z1 and z2.
(b) show that the angle between z and the x axis is the sum of the angles made by z1 and z2 separately.

3. The attempt at a solution
(a) i want to just do regular multiplication. (a+ib)(c+id)= ac-bd+i(ad+bc) however i dont see how that would show the length of z is the product of z1 and z2, all i did was multiply.

my next idea would be to take the magnitudes of z1 and z2 and multiply them. so, (√[a^2+(ib)^2]) * (√[c^2+(id)^2]) = (√[a^2-b^2])(√[c^2-d^2]).

(b)this would depend in part on which attempt of part (a) is correct. this is because depending on the correct way vector z is represented with its components the angle is going to be different.

2. Aug 27, 2013

### BrainHurts

I believe for part (a) you are trying to do

$|z| = |z_1||z_2|$ so you can verify that based on what you have you know that if $z = x + iy$ then $|z| = \sqrt{x^2 + y^2}$

(b) Use the fact that $z = r e^{i \theta}$ I think.

3. Aug 28, 2013

### vela

Staff Emeritus
Let $r_i = |z_i|$ and $\theta_i$ = angle between $z_i$ and the x-axis. Using basic trig, you should be able to see that you can write
\begin{align*}
a &= r_1\cos\theta_1 \\
b &= r_1\sin\theta_1 \\
c &= r_2\cos\theta_2 \\
d &= r_2\sin\theta_2
\end{align*} Try plugging that into your expression for the product.

You don't want to include the factor of $i$ when squaring. The magnitude of a complex number $z$ is given by $|z| = \sqrt{z z^*}$ where $z^*$ is the conjugate. If you work that out, you'll see the $i$ drops out.