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)
=  (x2 – y2) +  2x
Show
u(x,y) is a harmonic function
Solution:
If it’s harmonic then we must have:
¶ 2 u/ ¶ x2  +  ¶ 2
u/ ¶ y2     =  0
Take the 1st, 2nd partials:
¶ u/ ¶ x  = 2x + 2  
and  ¶ u/ ¶ y = -2y
  ¶ 2
u/ ¶ x2   =  2 
and  ¶ 2
u/ ¶ y2     =   =  -2
Therefore:  
¶ 2 u/ ¶ x2  +  ¶ 2
u/ ¶ y2     =   (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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