The Problems again:

1) (a) Given u(x,y) + iv(x,y) = 2x

^{2}+ i2y^{2}
find f(z,z*)

(b) Express f(z)= z

^{2}+ z – 3 in polar form

2) Solve for x and y:

**8x 2 + 3iy - 4 = 8y – 4iy**

3) Solve: (z+1) 3 = z 3

**Solutions:**

*(1)*Let: z = x + iy, and z* = x – iy

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

Then: y = (z – z*)/ 2i

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) z

^{2}= r^{2}exp(i2(q)) = r^{2}(cos (2q) + isin(2 q))
z = r exp(i(q)) = r(cos(q) + isin(q)

so: z

^{2 }+ z = r^{2}(cos (2q) + isin(2 q)) + r(cos(q) + isin(q)
Collecting like terms in i and simplifying:

f(z) = r

^{2}(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) = r

^{2 }(cos (2q) + r(cos(q)) – 3

3) 8x 2 + 3iy - 4 = 8y – 4iy

8x 2 = 4

So: x 2 = 4/ 8 = 1/ 2

Therefore: x =

**Ö**( 1/2) = 1/**Ö**2
=

**Ö**2**/**2
And:

8 y = - 3iy - 4iy = - 7 iy

y = -7 i/ 8

4) Expand the left side and set equal to the right:

z

^{3}+ 3z

^{2}+ 3z + 1 = z

^{3}

è3z

^{2}+3z + 1 = z

^{3 }-

^{ }z

^{3}

or 3z

^{2}+ 3z +1 = 0

(This can be solved using the quadratic formula, to give two roots)

Then: z1 = ½ + i Ö (3) / 6

Then: z1 = ½ + i Ö (3) / 6

And z2 = -(½ ) + iÖ (3)/ 6

Checking the result against the original equation:

z

z

^{3 }= 0.192i and (z^{3}+ 1)^{ 3}= 0.192i
So both quantities are equal, the roots are correct.

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