# Homework Help: How to take derivative of complex number?

1. Sep 1, 2009

### dumpman

1. The problem statement, all variables and given/known data
On the first day of Electromagnetism class, the professor gave this problem to us to solve. I never learn about taking derivative of complex number. Can someone give me some hints?
his problem was:
Given P= 0.5 Re(I*V)
I= V/(A+B)
A= R+jX , B=Y+ jZ

V is constant, find the derivative of P with respect to X (dP/dx)

2. Relevant equations

3. The attempt at a solution

2. Sep 1, 2009

### rock.freak667

You could simplify VI in terms of the complex variables, then take out the real part (which might just be a function in X only).

So that P=0.5 Re(I*V)= 0.5f(X)

3. Sep 1, 2009

### dumpman

thank you for your response, could you be more specific? which part is the real part and which is the imaginery part?

4. Sep 1, 2009

### rock.freak667

j or i is used to denote the complex variable √-1, so in A= R+jX, R is the real part (the one without a 'j')

5. Sep 1, 2009

### dumpman

this is what I did after your hint. VI becomes V^2/A+B, and V^2 is constant so I take it outside, leaving just 1/A+B for differentiation.
1/A+B = 1/R+jX+Y+jZ. now I am stuck. I dont know how to find dP/dX.

6. Sep 1, 2009

### rock.freak667

so 1/(A+B)= 1/(R+jX+Y+jZ)=1/[(R+Y)+(X+Z)j]

now multiply both the numerator and denominator by the conjugate of (R+Y)+(X+Z)j

7. Sep 1, 2009

### dumpman

thanks for your hint again, now I got [(R+Y)-j(X+Z)]/[(R+Y)^2+(X+Z)^2]

Last edited: Sep 1, 2009
8. Sep 1, 2009

### rock.freak667

so put that into the form a+jb and then the real part is simply your 'a'

9. Sep 1, 2009

### dumpman

so the real part is (R+Y)/[(R+Y)^2+(X+Z)^2]. can I take the derivative with respect with X? can I treat R,Y,Z as constant?

10. Sep 2, 2009

### rock.freak667

I am not sure what R,Y and Z are supposed to be, but I assume you would.

11. Sep 2, 2009

### dumpman

thank you again, after I worked it out, I got dP/dX = -0.5(V^2)[2(x+Z)]/(R+Y)[(R+Y)^2+(X+Z)^2]^2

12. Sep 2, 2009

### NJunJie

hi,

how to find the derivative of [1/(z*sin(z)*cos(z)] from first principles?
complicated. any recommendations?

13. Sep 2, 2009

### rock.freak667

Don't hijack this thread. Start a new one using the definition of the derivative.

$$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$