Question

If 4xy−kx2y−5y3 is the imaginary part of the analytic function f(z)=u+iv then, the value of k

Answer 1

By the Cauchy–Riemann equations we have $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=4x-kx^2-15y^2$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=-4y+2kxy$

Then $\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)=2ky$ and $\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)=-30y$. So $2k=-30$ and $k=-15$

Answer: $k=-15$