Taking e^z=e^x(cos y+isin y) (z=x+iy) as the definition…

Taking $e^z=e^x(\cos y+i\sin y)$ ($z=x+iy$) as the definition, prove: $\ 1^\circ\ e^{z_1}e^{z_2}=e^{z_1+z_2}$; $\ 2^\circ\ e^{z+2k\pi i}=e^z$ for $k\in\mathbb Z$; $\ 3^\circ$ if $e^{z+w}=e^z$ for all $z$, then $w=2k\pi i$ for some $k\in\mathbb Z$.

Part $1^\circ$ · $e^{z_1}e^{z_2}=e^{z_1+z_2}$

Let $z_1=x_1+iy_1,\ z_2=x_2+iy_2$. Then:

$$ e^{z_1}e^{z_2}=e^{x_1+x_2}(\cos y_1+i\sin y_1)(\cos y_2+i\sin y_2) $$

Expanding the product and using the angle-addition formulas $\cos(y_1+y_2)=\cos y_1\cos y_2-\sin y_1\sin y_2$, $\sin(y_1+y_2)=\sin y_1\cos y_2+\cos y_1\sin y_2$:

$$ e^{z_1}e^{z_2}=e^{x_1+x_2}[\cos(y_1+y_2)+i\sin(y_1+y_2)]=e^{(x_1+x_2)+i(y_1+y_2)}=e^{z_1+z_2} \quad\checkmark $$

Part $2^\circ$ · $e^{z+2k\pi i}=e^z$

$z+2k\pi i=x+i(y+2k\pi)$, so:

$$ e^{z+2k\pi i}=e^x[\cos(y+2k\pi)+i\sin(y+2k\pi)]=e^x(\cos y+i\sin y)=e^z $$

using the $2\pi$-periodicity of $\sin,\cos$ for any integer $k$.

Part $3^\circ$ · The only period is $2k\pi i$

Step 1 · Reduce to $e^w=1$

By Part $1^\circ$: $\ e^{z+w}=e^z\Longleftrightarrow e^ze^w=e^z\Longleftrightarrow e^w=1\ $ (dividing by $e^z\neq0$), and this must hold independent of $z$, so the condition is exactly $e^w=1$.

Step 2 · Split into modulus and argument

Write $w=a+ib$. Then $e^w=e^a(\cos b+i\sin b)=1$ means:

$$ e^a\cos b=1, \qquad e^a\sin b=0 $$

Step 3 · Solve

Since $e^a>0$, the second equation forces $\sin b=0$, i.e. $b=k\pi$ for some $k\in\mathbb Z$. Substituting into the first: $\ e^a\cos(k\pi)=e^a(-1)^k=1$. Since $e^a>0$, we need $(-1)^k=1$, i.e. $k$ even, say $k=2m$; then $e^a=1\Rightarrow a=0$.

Step 4 · Conclusion

$$ w=0+i(2m\pi)=2m\pi i, \qquad m\in\mathbb Z $$

$$ \boxed{e^{z+w}=e^z \text{ for all } z \ \Longleftrightarrow\ w=2k\pi i,\ k\in\mathbb Z} $$

Verification

$w=2\pi i$: $e^{2\pi i}=\cos2\pi+i\sin2\pi=1$ ✓. $w=\pi i$ (a value of the excluded odd-multiple form, $k=\frac12\notin\mathbb Z$): $e^{\pi i}=-1\neq1$, correctly excluded.