Solutions to Complex Functions Problems

Solutions:

(1)   Let:  z = x + iy, and z* = x – iy

Adding:   z + z* = (x + iy) + (x - iy) = 2x

We see: x = (z + z*)/2

Subtracting:

(z – z*) = [x + iy – x + iy] =  i2y

Then:  y = (z – z*)/ 2i

We can now formulate the function f(z,z*):

f(z,z*) = 2[(z + z*)/2]²  + i2[(z – z*)/ 2i]²

2)  Rewrite:  z² + z – 3   in polar form:

z² = r² exp(i2(q)) = r² (cos (2q) + isin(2 q))

z = r exp(i(q)) = r(cos(q) + isin(q)

so:

 z² + z = r² (cos (2q) + isin(2 q)) +  r(cos(q) + isin(q)

Collecting like terms in i and simplifying:

f(z) =  r²(cos (2q) + r(cos(q)) + i{sin(2q) + sin(q)} – 3

so:

 iv(r, q) =  i{sin(2q) + sin(q)}

and 

v (r, q) =  {sin(2q) + sin(q)}

while:

u(r, q) =  r² (cos (2q) + r(cos(q)) – 3

3)     8x ² + 3iy  -  4    = 8y   – 4iy

 Write real function as:

 8x ²    =   4

So:   x ²    =   4/ 8   =  1/ 2

Therefore:  x =  Ö( 1/2)  =   1/ Ö2

=  Ö2/ 2

And for complex part:  

8 y   =   - 3iy  - 4iy   =   - 7 iy

y  =   -7 i/  8