Question - Solving Complex Number Division
Solution:
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 $$ \frac{z_1}{z_2} $$.
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