Discussion
\(F(z)\) is a function of the complex variable \(z = x + iy\) given by
\[ F(z) = i\,z + k\,Re(z) + i\,Im(z). \]
For what value of \(k\) will \(F(z)\) satisfy the Cauchy-Riemann equations?
Answer: B
The required value is k = 1.
Substitute z = x + iy:
\[ \begin{aligned} F &= i(x + iy) + kx + iy \\ &= (kx - y) + i(x + y) \end{aligned} \]
So u = kx - y and v = x + y. The Cauchy-Riemann equations need \(u_{x} = v_{y}\), which gives k = 1, and \(u_{y} = -v_{x}\), which gives -1 = -1 and holds for any k.
k = 0 or -1 breaks the first equation, and k = y is not a constant; with k = 1, F(z) = (1 + i)z, which is plainly analytic.
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