Using the definitions of the corresponding functions, prove…
Using the definitions of the corresponding functions, prove: $\ \cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$.
Explanation of Method
Apply $\cos z=\frac{e^{iz}+e^{-iz}}2$ directly to $z=x+iy$, use $i^2=-1$ to separate the real exponential factors, then invoke Euler's formula and regroup into $\cosh y,\sinh y$.
Step 1 · Apply the definition
$$ \cos(x+iy)=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}2=\frac{e^{ix-y}+e^{-ix+y}}2=\frac{e^{-y}e^{ix}+e^{y}e^{-ix}}2 $$
Step 2 · Apply Euler's formula
$$ =\frac{e^{-y}(\cos x+i\sin x)+e^{y}(\cos x-i\sin x)}2=\frac{\cos x(e^{y}+e^{-y})-i\sin x(e^{y}-e^{-y})}2 $$
Step 3 · Recognize hyperbolic functions
$$ \cos(x+iy)=\cos x\cdot\frac{e^y+e^{-y}}2-i\sin x\cdot\frac{e^y-e^{-y}}2=\cos x\cosh y-i\sin x\sinh y $$
Checks at special cases
$y=0$: $\ \cos x\cosh0-i\sin x\sinh0=\cos x\cdot1-0=\cos x$ ✓. $\ x=0$: $\ \cos0\cosh y-i\sin0\sinh y=\cosh y$, confirming $\cos(iy)=\cosh y$ ✓.
Result
Real part $\text{Re}[\cos z]=\cos x\cosh y$, imaginary part $\text{Im}[\cos z]=-\sin x\sinh y$ — this is the workhorse formula used throughout for splitting $\cos z$ into real and imaginary parts.