$\overline{\sin(z)} = \sin(\overline{z})$ and $\overline{\cos(z)} = \cos(\overline{z})$
by Hidenori
Proposition
Prove that $\overline{\sin(z)} = \sin(\overline{z})$ and $\overline{\cos(z)} = \cos(\overline{z})$”
Solution
\(\begin{align*} \overline{\exp(z)} &= \overline{\exp(x + iy)} \\ &= \overline{\exp(x)\exp(iy)} \\ &= \overline{\exp(x)(\cos(y) + i\sin(y))} \\ &= \overline{\exp(x)\cos(y) + i\exp(x)\sin(y))} \\ &= \exp(x)\cos(y) - i\exp(x)\sin(y) \\ &= \exp(x)(\cos(y) - i\sin(y)) \\ &= \exp(x)\exp(-iy) \\ &= \exp(x-iy) \\ &= \exp(\overline{z}). \end{align*}\)
for any $z \in \mathbb{C}$.
Therefore,
\[\begin{align*} \overline{\sin(z)} &= \overline{\Big(\frac{\exp(iz) - \exp(-iz)}{2i}\Big)} \\ &= \frac{\overline{\exp(iz)} - \overline{\exp(-iz)}}{-2i} \\ &= \frac{\exp(-i\overline{z}) - \exp(i\overline{z})}{-2i} \\ &= \frac{\exp(i\overline{z}) - \exp(-i\overline{z})}{2i} \\ &= \sin(\overline{z}). \end{align*}\]Similarly,
\[\begin{align*} \overline{\cos(z)} &= \overline{\Big(\frac{\exp(iz) + \exp(-iz)}{2}\Big)} \\ &= \frac{\overline{\exp(iz)} + \overline{\exp(-iz)}}{2} \\ &= \frac{\exp(-i\overline{z}) + \exp(i\overline{z})}{2} \\ &= \frac{\exp(i\overline{z}) + \exp(-i\overline{z})}{2} \\ &= \cos(\overline{z}). \end{align*}\]Subscribe via RSS