Consider the following expression:

z = sin (y + it) + cos (y - it)

where z, y, and t are variables, and i = √-1 is a complex number. The partial differential equation derived from the above expression is

Explanation-

Given equation is, z = sin (y + it) + cos (y - it)

Partially differentiating the equation with respect to y, we have 

\(\frac{{\delta z}}{{\delta y}} = \)cos (y + it) - sin (y - it)

Partially differentiating the equation with respect to y twice, we have

\(\frac{{{\partial ^2}z}}{{\partial {y^2}}} = \) -sin (y + it) - cos (y - it)---------(i)

Partially differentiating the equation with respect to t, we have 

\(\frac{{\delta z}}{{\delta t}} = \) i cos (y + it) + i sin (y - it)

Partially differentiating the equation with respect to y twice, we have

\(\frac{{{\partial ^2}z}}{{\partial {t^2}}} = \) i² sin (y + it) - i² cos (y - it)-------(ii)

Value of i² = -1

So equation (ii) will be, 

\(\frac{{{\partial ^2}z}}{{\partial {t^2}}} = \) cos (y - it) + sin (y + it)------(iii)

\(\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=\) -sin (y + it) - cos (y - it)+cos (y - it) + sin (y + it) = 0

So the partial differential equation derived from the above expression is \(\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=0\).