How do the lengths and angles of vectors relate in complex multiplication?

bfusco
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Homework Statement


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.

The Attempt at a Solution


(a) i want to just do regular multiplication. (a+ib)(c+id)= ac-bd+i(ad+bc) however i don't 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.

Thank you in advance.
 
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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.
 
bfusco said:

Homework Statement


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.

The Attempt at a Solution


(a) i want to just do regular multiplication. (a+ib)(c+id)= ac-bd+i(ad+bc) however i don't see how that would show the length of z is the product of z1 and z2, all i did was multiply.
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.

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]).
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.

(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.

Thank you in advance.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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