Solving Complex Number Division
Certainly! To find \( \frac{z_1}{z_2} \) given that \( z_1 = \cos \theta + j \sin \theta \) and \( z_2 = \cos \theta - j \sin \theta \), we can directly divide the complex numbers as follows:
\[ \frac{z_1}{z_2} = \frac{\cos \theta + j \sin \theta}{\cos \theta - j \sin \theta} \]
To divide these two complex numbers, we can multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary parts from the denominator. The conjugate of \( z_2 \) is \( \cos \theta + j \sin \theta \), which is actually the same as \( z_1 \).
So, we multiply both top and bottom by this conjugate:
\[ \frac{z_1}{z_2} = \frac{(\cos \theta + j \sin \theta)(\cos \theta + j \sin \theta)}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)} \]
Multiplying out the numerator, we get:
Numerator: \( (\cos^2 \theta + 2j \cos \theta \sin \theta - \sin^2 \theta) \)
(using \( j^2 = -1 \))
Multiplying out the denominator, we get:
Denominator: \( (\cos^2 \theta - (\sin^2 \theta)) \)
Simplify the denominator using \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \):
Denominator: \( \cos(2\theta) \)
However, the denominator simplifies even further because \( \cos^2 \theta - \sin^2 \theta = 1 \) (from the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \)).
Therefore, the denominator actually is:
Denominator: \( 1 \)
Returning to the numerator, we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) again:
Numerator: \( 1 + 2j \cos \theta \sin \theta \)
The numerator simplifies using the double angle identity for sine: \( 2\sin \theta \cos \theta = \sin(2\theta) \):
Numerator: \( 1 + j \sin(2\theta) \)
Putting it all together, we have:
\[ \frac{z_1}{z_2} = 1 + j \sin(2\theta) \]
This is the final simplified form of \( \frac{z_1}{z_2} \).
