Let \(s,t:\mathbb R\to\mathbb R\) be differentiable and suppose \[ f(x+iy)=e^x\bigl(s(y)+it(y)\bigr) \] is holomorphic on \(\mathbb C\), with \[ s(0)=1, \qquad t(0)=0. \] Determine \(s(y)\) and \(t(y)\).
Solution
Write \[ f(x+iy)=u(x,y)+iv(x,y), \] where \[ u(x,y)=e^x s(y), \qquad v(x,y)=e^x t(y). \]
1. The Cauchy–Riemann equations imply\[ s'(y)=-t(y), \qquad t'(y)=s(y) \]for every \(y\in\RR\).
Since \(f\) is holomorphic, \[ u_x=v_y, \qquad u_y=-v_x. \] The four partial derivatives are \[ u_x=e^x s(y), \qquad u_y=e^x s'(y), \] and \[ v_x=e^x t(y), \qquad v_y=e^x t'(y). \] Because \(e^x>0\), the first Cauchy–Riemann equation gives \(s=t'\), while the second gives \(s'=-t\).
2. Define\[ g(y)=s(y)+it(y). \]Then\[ g'(y)=ig(y) \]and\[ g(0)=1. \]
By step 1, \[ \begin{aligned} g'(y) &=s'(y)+it'(y)\\ &=-t(y)+is(y)\\ &=i\bigl(s(y)+it(y)\bigr)\\ &=ig(y). \end{aligned} \] The initial conditions give \[ g(0)=s(0)+it(0)=1. \]
3. One has\[ g(y)=e^{iy} \]for every \(y\in\RR\).
Set \[ h(y)=e^{-iy}g(y). \] By step 2, \[ \begin{aligned} h'(y) &=-ie^{-iy}g(y)+e^{-iy}g'(y)\\ &=e^{-iy}\bigl(-ig(y)+ig(y)\bigr)\\ &=0. \end{aligned} \] Hence \(h\) is constant. Since \[ h(0)=g(0)=1, \] we have \(h(y)=1\), and therefore \(g(y)=e^{iy}\).
4. The required functions are\[ \boxed{s(y)=\cos y}, \qquad \boxed{t(y)=\sin y}. \]
By step 3 and Euler’s formula, \[ s(y)+it(y) = g(y) = e^{iy} = \cos y+i\sin y. \] Equality of real and imaginary parts gives the result.
5. Q.E.D.
Step 4 determines both functions.