Working backwards in Complex Analysis.. how to get xy+1 + i/2(y^2-x^2) = -i/2(z^2 +2i)?

Question

xy+1 + i/2(y^2-x^2) = -i/2(z^2 +2i)

I can see how it is true (i.e if I put z=x+iy into the answer, i'd get the first bit) but I just can't think how I would work in reverse so I would know how to put z^2+2i inside the brackets...

Gee I'm getting tired now, I dont think my questions are making sense.... I have a baby that doesn't sleep and an exam in 11 days and counting... thanks for any input

Answer 1

xy+1 + i/2(y^2-x^2) = -i/2(z^2 +2i)

x=rcosa
y=rsina

xy=r^2sinacosa
=1/2*r^2 sin2a
x^2+y^2=(rcosa)^2+(rsina)^2=1
y^2-x^2=(rsina)^2-(rcosa)^2= - r^2 (cos2a)

1+1/2*r^2 sin2a+i/2(- r^2 (cos2a))= -i/2(z^2 +2i)

Answer 2

First : U have to state that x,y are real & z is complex ?
Second : is the given expression in the form :
xy+1 + i/[2(y^2-x^2)] = -i/[2(z^2 +2i)] or
xy+1 + (i/2)(y^2-x^2) = (-i/2)(z^2 +2i) to be more accurate
If it is the seond form then :
Multiplying both sides by 2i
2i xy + 2i - y^2 + x^2 = z^2 + 2i
So : z^2 = x^2 + 2i xy - y^2 = x^2 + 2i xy +i^2y^2 = (x+iy)^2
So z= x+ iy
or z = -x - iy

Answer 3

As a general thing, if you are given the real and imaginary parts w = u(x,y) + i*v(x,y) of some complex-valued function, you are not necessarily going to be able to find a simple function f such that w = f(z) where z=x+i*y. The technique of replacing x with r*cos(a), etc. works here because there is in fact a simple answer.

There are certain tests which can be applied to u and v (the Cauchy-Riemann equations) which tell you whether or not the function f is analytic. Some analytic functions have nice expressions in terms of elementary functions (others are only expressible as power series), but if u and v do not satisfy the CR equations, then it is very unlikely that you will get a satisfactory w=f(z) expression. If they do satisfy the equations, it is worth trying to find f (you may not succeed however).

Answer 4

If z = x + iy, denote z conjugate z* = x - iy.
Then x = (z + z*)/2, y = (z - z*)/2i.
which gives:
y^2 - x^2 = - (z^2 - 2zz* + z*^2)/4 - (z^2 + 2zz* + z*^2)/4
= - (z^2 + z*^2)/2...........(1), and
xz = (z^2 - z*^2)/4i.........(2)
Plugging (1) and (2) into your formula:
xy + 1 + i/2(y^2 - x^2)
= (z^2 - z*^2)/4i + 1 - i/2(z^2 + z*^2)/2
= -i/2((z^2 - z*^2)/2 + (z^2 + z*^2)/2 + 2i)
= -i/2(z^2 + 2i).

Answer 5

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