The Problems again:
1) (a) Given u(x,y) + iv(x,y) = 2x2 + i2y2
find f(z,z*)
(b) Express f(z)= z2 + 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) 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
8x 2 = 4
So: x 2 = 4/ 8 = 1/ 2
Therefore: x = Ö ( 1/2) = 1/ Ö2
= 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 3 = 0.192i and (z 3 + 1) 3 = 0.192i
z 3 = 0.192i and (z 3 + 1) 3 = 0.192i
So both quantities are equal, the roots are correct.
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