1) f(z) = u(x,y) + iv(x,y) =
cos x cosh y – i(sinx sinh y)

We have to verify the Cauchy- Riemann equations. We note
that:

u(x,y) = cos x cosh
y and v(x,y) = - sinx sinh y

Then:

¶ u/ ¶ x = - sinx cosh y = ¶ v/ ¶ y

And:

¶ v/ ¶ x = - cos x sinh y = - ¶ u/ ¶ y

So the Cauchy_Riemann equations are satisfied

b2) Let
u(x,y)
= (x

^{2 }– y^{2}) + 2x
Show
u(x,y) is a harmonic function

Solution:
If it’s harmonic then we must have:

¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2 }= 0
Take the 1st, 2nd partials:

¶ u/ ¶ x = 2x + 2
and ¶ u/ ¶ y = -2y

¶

^{2}u/ ¶ x^{2}= 2 and ¶^{2}u/ ¶ y^{2 }= = -2
Therefore:

¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2 }= (2) + (-2) = 0*SO that u(x,y) is a harmonic function*.

b) Hence or otherwise find the harmonic conjugate v(x,y)

We’ve already shown:

¶ u/ ¶ x = 2x + 2
and ¶ u/ ¶ y = -2y

By the Cauchy –Riemann equations:

¶ v/ ¶ x =
- ¶ u/ ¶ y = 2y

And:

¶ u/ ¶ x = ¶ v/ ¶ y = 2x + 2

Take the differential using the chain rule:

dv = (¶ v/ ¶ x ) dx +
(¶ v/ ¶ y ) dy

Substitute from Cauchy-Riemann equations:

dv = 2y (dx) + (2x + 2) dy

Integrating:

v = 2 xy + 2xy + 2y

Or:

v = 4xy + 2y = 2y (2x +
1) + C

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