・$f(z)=f(x+iy)=e^xu(y)+ie^xv(y)$.
・$f(z)$ is regular.
・$f(x)=e^x.$ (When $y=0$, $f(z)=e^x$.)
Under this condition, determine $u(y), v(y).$
The answer is following.
From Cauchy–Riemann equations, $u(y)=v'(y), u'(y)=-v(y).$
So, $(u(y)^2+v(y)^2)'=2u(y)u'(y)+2v(y)v'(y)=0.$
Thus $u(y)^2+v(y)^2=k$, and from $f(x)=e^x,$ $u(0)=1, v(0)=0.$ Then $k=1.$
Therefore, $u(y)^2+v(y)^2=1$.
The answer is $u(y)=\cos y, v(y)=\sin y.$ ■
But I'm not convinced. This answer says that $u(y)^2+v(y)^2=1 \Longrightarrow u(y)=\cos y, v(y)=\sin y.$ Aren't there other $u(y),v(y)$ ? Only $u(y)=\cos y, v(y)=\sin y$ is solution?
I'm not convinced either. But the answer is correct. Note that $u''=-v'=-u$ and therefore $u(x)=\alpha\cos(x)+\beta\sin(x)$ for some constants $\alpha$ and $\beta$, and then $v=-u'=\alpha\sin(x)-\beta\cos(x)$. But $\alpha=u(0)=1$ and $\beta=v(0)=0$.