f(x + iy) = u + iv

and since u,v depend on x and y, they can be considered as real functions of the real variables x and y such that:

u = u(x,y) and v = v(x,y)

Example: write f(z) = z

^{2}in the form f(z) = u(x,y) + iv(x,y)

If z = (x + iy) then: z

^{2}= (x + iy)

^{2}= x

^{2}+ i2xy - y

^{2}

= (x

^{2 }– y^{2}) + i2xy
The last step above shows how the complex function is separated into two parts, one with the factor i, the other without. The one with the factor applies to the function v(x,y) so:

v(x,y) = 2xy

While: u(x,y) = x

^{2}+ y^{2}
Conversely, of course, one can be given the functions u(x,y) and v(x,y) then be asked to find f(z), e.g. in terms of z and-or its complex conjugate,

**.***z**
Example:

Given u(x,y) + iv(x,y) = = 4x

^{2}+ i4y^{2}
Find f(z,z*)

Again, let: z = x + iy, and z* = x – iy

Adding:

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

So we see: x = (z + z*)/2

Now,

*subtracting*: (z – z*) = [x + iy – x + iy] = i2y
So: y = (z – z*)/ 2i

Since we have both x and y we can now formulate the function f(z,z*):

f(z,z*) = 4[(z + z*)/2]

^{2 }+ i4[(z – z*)/ 2i]^{2}
Before leaving the basics of complex functions, it’s important to note that a polar form is also used, viz.

f (z) = f(r exp(i(q)) = u(r, q) + iv(r, q)

Example: Express f(z)= z

^{2}in polar form.
z

^{2}= r^{2}exp(i2(q)) = r^{2}(cos (2q) + isin(2 q))
Therefore:

u(r, q) = r

^{2}(cos (2q) and:
v (r, q) = r

^{2}(sin (2q))
Problems for the Math Maven:

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

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

^{2}+ z – 3 in polar form
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