Proving $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}\Re{(\cosh z}) = \cosh(\Re z)\cdot \cos(\Im z)$

Question

I want to show $\Re{(\cosh z)}) = \cosh(\Re z)\cdot \cos(\Im z)$.

I did the following:

\begin{align*} \cosh(\Re z)\cdot \cos(\Im z) &= \frac {1}{2}(e^{\Re z}+e^{-\Re z})\cdot \frac {1}{2}(e^{i\Im z}+e^{-i\Im z}) \\& = \frac {1}{4}(e^{\Re z}+e^{-\Re z})\cdot (e^{i\Im z}+e^{-i\Im z}) \\ &= \frac {1}{4}(e^{\Re z}+e^{-\Re z}) \cdot 2\\ & = \frac {1}{2}(e^{\Re z}+e^{-\Re z}) \\ &= \frac {1}{2}\Re(e^z+e^{-z}) \\ &= \Re(\cosh z) \end{align*}

Using: $\cosh(x) = \frac {1}{2}(e^x+e^{-x})$ and $\cos(x) = \frac {1}{2}(e^{ix}+e^{-ix})$.

I would be very glad if someone could tell me if this is correct!

Answer 1

As an alternative, but taking the path of OP so that his /her error can be seen: $$ \cosh\left[\operatorname{Re}(z)\right] \cdot \cos\left[\operatorname{Im}(z)\right] = \left(\frac{e^{\operatorname{Re}(z)} + e^{-\operatorname{Re}(z)}}{2}\right) \left(\frac{e^{i\operatorname{Im}(z)} + e^{-i\operatorname{Im}(z)}}{2}\right)$$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)}}{4} $$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + }{4} $$ $$ = \frac{e^{Re(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} $$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) + i\operatorname{Im}(z)]}}{4} + \frac{e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} = \frac{\cosh {z}}{2} + \frac{\cosh {\overline {z}}}{2}$$ $$ = \frac{\cosh {z} + \cosh {\overline {z}}}{2} = \frac{2\operatorname{Re}{\left[\cosh{z}\right]}}{2} = \operatorname{Re}{\left[\cosh{z}\right]}$$

Answer 2

Let $z=x+iy$, then we have that

$$\cosh z=\frac{e^z+e^{-z}}2=\frac12e^xe^{iy}+\frac12e^{-x}e^{-iy}=\frac12e^x(\cos y+i\sin y)+\frac12e^{-x}(\cos y-i\sin y) $$

$$ \implies \operatorname{Re}(\cosh z)=\frac12e^x\cos y+\frac12e^{-x}\cos y=\frac{e^x+e^{-x}}2\cos y=\cosh x \cdot\cos y$$

Answer 3

Using the sum of angles formulae,

$\cosh(x+iy)=\cosh(x)\cosh(iy)+\sinh(x)\sinh(iy)$

but $ \cosh(iy)=\cos(y)$ and $\sinh(iy)=i\sin(y)$

so: $\cosh(x+iy)=\cosh(x)\cos(y) + i \sinh(x)\sin(y)$

QED