Question

If f(z)=u+iv is analytic and u-v=ex(cos y-sin y) find f(z) in terms of z.

Answer 1

If
`f(z)=f(x+iy)=u(x,y)+iv(x,y)` is analytic, then the Cauchy-Riemann conditions hold:


`\begin{cases}(\partial u)/(\partial x)=(\partial v)/(\partial y)\\\\(\partial u)/(\partial y)=-(\partial v)/(\partial x)\end{cases}`

I'll denote these partial derivatives by
`u_x,v_y,u_y,v_x`.

You have


`\begin{cases}(u-v)_x=u_x-v_x=e^x(\cos y-\sin y)&(1)\\(u-v)_y=u_y-v_y=e^x(-\sin y-\cos y)&(2)\end{cases}`

Adding and subtracting
`(1)` and
`(2)`, respectively, gives


`\begin{cases}(u_x-v_y)+(u_y-v_x)=-2e^x\sin y&(1)+(2)\\(u_x+v_y)-(u_y+v_x)=2e^x\cos y&(1)-(2)\end{cases}`

Because
`f(z)` is analytic, this system reduces to


`\begin{cases}2u_y=-2e^x\sin y&(3)\\2u_x=2e^x\cos y&(4)\end{cases}\iff\begin{cases}u_y=-e^x\sin y&(3)\\u_x=e^x\cos y&(4)\end{cases}`

Integrating
`(3)` with respect to
`y` gives


`\displaystyle\int u_y\,\mathrm dy=\int-e^x\sin y\,\mathrm dy`

`u=e^x\cos y+g(x)`

Differentiating with respect to
`x` gives


`u_x=e^x\cos y+g'(x)=e^x\cos y\implies g'(x)=0\implies g(x)=C_1`

where
`C_1` is an arbitrary constant.

On the other hand, you have


`\begin{cases}-v_x=-e^x\sin y&(5)\\v_y=e^x\cos y&(6)\end{cases}`

(Note that
`(5)=(3)` and
`(6)=(4)` by the C-R conditions.) Integrating
`(5)` with respect to
`x` gives


`\displaystyle\int-v_x\,\mathrm dx=\int-e^x\sin y\,\mathrm dx`

`v=e^x\sin y+h(y)`

Differentiating with respect to
`y` gives


`v_y=e^x\cos y+h'(y)=e^x\cos y\implies h'(y)=0\implies h(y)=C_2`

It follows that


`f(x+iy)=(e^x\cos y+C_1)+i(e^x\sin y+C_2)`

`f(x+iy)=e^x(\cos y+i\sin y)+(C_1+iC_2)`

`f(z)=e^(iz)+C_3`

where the last equality is an invocation of Euler's formula.