# Derivate of complex numbers

1. Nov 8, 2009

### soopo

1. The problem statement, all variables and given/known data
Why do you need to use Cauchy-Riemann in the derivate of $x^3 + i(1-y)^3$, while not in the derivate of $\frac { 1 } {z^2 + 1}$ in using the quotient rule?

where z = x + iy.

I derivated the first expression implicitly in the exam which resulted in zero points of the exercise.

3. The attempt at a solution

I would derivate the latter by expanding the denominator such that

$$\frac{1} { (x+iy)^2 +1} =...= \frac {x^2 - y^2 +1} {x^4 + y^4 - 6x^2 y^2 +1} - \frac {2xyi} {x^4 + y^4 -6x^2 y^2 +1}$$

and then derivate as real and then as complex.
The expression is rather challenging at the moment.
I am not sure what is the best way to derivate the latter statement correctly.

Last edited: Nov 8, 2009
2. Nov 8, 2009

### sara_87

I'm guessing that to get:
$$\frac{1}{ (x+iy)^2 +1} =...= \frac{x^2 - y^2 +1}{x^4 + y^4 - 6x^2 y^2 +1} - \frac{2xy}{x^4 + y^4 -6x^2 y^2 +1}$$

you multiplied by the conjugate, so you multiplied top and bottom by:
[x^{2}-y^{2}+1]-2xyi
right?

if that's right, i dont agree with what you got. can you please show one more intermediate step?

3. Nov 8, 2009

### soopo

I did multiply by the conjugate.

The denominator is

$$(x + iy)^2 +1 = x^2 + 2xyi -y^2 +1$$
$$= (x^2 - y^2 +1) + 2xyi$$

Multiplying by the conjugate

$$\frac{1} { (x^2 - y^2 +1) + 2xyi } = \frac { (x^2 - y^2 +1) - 2xyi } { (x^2 - y^2 +1)^2 + (2xy)^2 }$$

Let $A = x^2 - y^2$. Considering the denominator
$$(A +1)^2 = A^2 +2A +1$$
$$= (x^2 - y^2)^2 + 2(x^2 - y^2) +1$$
$$= x^4 - 2x^2 y^2 + y^4 + 2x^2 - 2y^2 +1$$

Making the projection: $B |-> B + (2xy)^2$, we get for the denominator

$$x^4 + y^4 + 2x^2 -2y^2 +1$$,

which is the same denominator as in my question.

The numerator is simply the conjugate $(x^2 - y^2 +1) - 2xyi$.

4. Nov 8, 2009

### sara_87

you mean, making the projection;
we get for the denominator:

$$B+(2xy)^{2}=x^4-2x^2y^2+y^4+2x^2-2y^2+1+4x^2y^2=x^4+2x^2y^2+y^4+2x^2-2y^2+1$$

?

5. Nov 8, 2009

### soopo

Yes, I mean what you say.
I did not use apparently the term "projection" correctly.
Perhaps, a more appropriate term would be a mapping or simply a plus between the two.