Question 3.8: Find f(z) in Problem 3.7.

Method 1

We have f(z) = f(x + iy) = u(x, y) + iv(x, y) .

Putting y = 0          f(x) = u(x, 0) + iv(x, 0).

Replacing x by z,      f(z) = u(z, 0) + iv(z, 0).

Then, from Problem 3.7, u(z, 0) = 0, v(z, 0)=z e^{-z} and so f(z)=u(z, 0)+i v(z, 0)=i z e^{-z}, apart from an arbitrary additive constant.

Method 2

Apart from an arbitrary additive constant, we have from the results of Problem 3.7,

Method 3

We have x=(z+\bar{z}) / 2, y=(z-\bar{z}) / 2 i. Then, substituting into u(x, y)+i v(x, y), we find after much tedious labor that \bar{z} disappears and we are left with the result i z e^{-z}.

In general, method 1 is preferable over methods 2 and 3 when both $u$ and v are known. If only u (or v ) is known, another procedure is given in Problem 3.101.