then prove that $ \dfrac{\partial^2u}{\partial \theta^2} + \dfrac{\partial^2u}{\partial \phi^2} \;=\; 4xy \dfrac{\partial^2 u}{\partial x \partial y} $

x$ +y=2e^{\theta} cos\phi \; \; x - y \;=\; 2ie^{\theta} sin\phi $

$ \\ $

Adding given equations, $ \\ \; \\ \therefore 2x \;=\; 2e^{\theta} cos\phi \; \; + 2ie^{\theta} sin\phi \\ \; \\ \therefore x \;=\; e^{\theta} cos\phi \; \; + ie^{\theta} sin\phi \;=\; e^{\theta} \cdot e^{i\theta} \;=\; …