Find f(z) from u - v = eˣ(cos y - sin y)

Short answer

Using (1+i)f(z), U=u−v=eˣ(cos y−sin y) is harmonic; then f(z)=eᶴ+c.

If f(z) = u + iv is any analytic function of complex variable z and u - v = eˣ(cos y - sin y), find f(z) in terms of z.

Given u − v = eˣ (cos y − sin y).

Form the analytic function F(z) = (1 + i) f(z):

(1 + i)(u + iv) = (u − v) + i(u + v) = U + iV

where U = u − v = eˣ(cos y − sin y) and V = u + v. Since F is analytic, U is harmonic and we can find F by the Milne-Thomson method.

U_x = eˣ (cos y − sin y)
U_y = eˣ (−sin y − cos y)

F′(z) = U_x − i U_y. Put x = z, y = 0:

F′(z) = e^z (1 − 0) − i e^z (−0 − 1) = e^z + i e^z = (1 + i) e^z

Integrate: F(z) = (1 + i) e^z + c₁.

f(z) = F(z)/(1 + i) = e^z + c, where c = c₁/(1 + i) is a constant.

For f = e^z = eˣ cos y + i eˣ sin y: u = eˣ cos y, v = eˣ sin y, so u − v = eˣ(cos y − sin y) ✓.

(The constant must make u − v unchanged, so c = β(1 + i) with β real; the simplest choice is c = 0.)

Answer: f(z) = e^z + c

✓ Answer checked by Claude Opus 5.5 (high effort)

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