Question

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Answer 1

Answer on Question #55954 - Math – Complex Analysis

If $z = \cos \theta + j \sin \theta$, prove that when $n$ is a natural number

$$\sin n \theta = \frac {1}{2 j} \left(z ^ {n} - \frac {1}{z ^ {n}}\right).$$

Solution

$$z = \cos \theta + j \sin \theta = e ^ {j \theta}$$

According to Euler's formula:

$$e ^ {j n \theta} = \cos (\boldsymbol {n} \boldsymbol {\theta}) + j \sin (\boldsymbol {n} \boldsymbol {\theta})$$ $$e ^ {- j n \theta} = \cos (- \boldsymbol {n} \boldsymbol {\theta}) + j \sin (- \boldsymbol {n} \boldsymbol {\theta}) = \cos (\boldsymbol {n} \boldsymbol {\theta}) - j \sin (\boldsymbol {n} \boldsymbol {\theta}).$$

Subtracting the second formula from the first one and dividing by $(2j)$ obtain

$$\sin (\boldsymbol {n} \boldsymbol {\theta}) = \frac {e ^ {j n \theta} - e ^ {- j n \theta}}{2 j}.$$

On the other hand,

$$e ^ {j n \theta} - e ^ {- j n \theta} = \left(e ^ {j \theta}\right) ^ {n} - \left(e ^ {j \theta}\right) ^ {- n} = z ^ {n} - z ^ {- n} = z ^ {n} - \frac {1}{z ^ {n}}$$

Thus,

$$\sin \boldsymbol {n} \boldsymbol {\theta} = \frac {1}{2 j} \left(\boldsymbol {z} ^ {n} - \frac {1}{\boldsymbol {z} ^ {n}}\right).$$

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