How to take derivative of complex number?

In summary: (x+h)^2\rightarrow\lim_{h \rightarrow 0} \frac{f(x+h)+f(x+2h)}{h}(x+h)^3\rightarrow\lim_{h \rightarrow 0} \frac{f(x+3h)+f(x+4h)}{h}(x+h)^4\rightarrow\lim_{h \rightarrow 0} \frac{f(x+5h)+f(x+6h)}{h}(x+h)^5\rightarrow\lim_{h \rightarrow 0} \frac{f(x+7h)+f(x+8
  • #1
dumpman
17
0

Homework Statement


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)



Homework Equations





The Attempt at a Solution

 
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  • #2
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
thank you for your response, could you be more specific? which part is the real part and which is the imaginery part?
 
  • #4
dumpman said:
thank you for your response, could you be more specific? which part is the real part and which is the imaginery part?

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
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 don't know how to find dP/dX.
 
  • #6
dumpman said:
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 don't know how to find dP/dX.

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
thanks for your hint again, now I got [(R+Y)-j(X+Z)]/[(R+Y)^2+(X+Z)^2]
 
Last edited:
  • #8
dumpman said:
thanks for your hint again, now I got [(R+Y)-j(X+Z)]/[(R+Y)^2+(X+Y)^2]

so put that into the form a+jb and then the real part is simply your 'a'
 
  • #9
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
dumpman said:
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?

I am not sure what R,Y and Z are supposed to be, but I assume you would.
 
  • #11
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
hi,

how to find the derivative of [1/(z*sin(z)*cos(z)] from first principles?
complicated. any recommendations?
 
  • #13
NJunJie said:
hi,

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

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

[tex]f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/tex]
 

Related to How to take derivative of complex number?

1. What is the definition of a derivative of a complex number?

The derivative of a complex number is a mathematical concept that describes the rate of change of the complex number with respect to another variable. It is equivalent to the slope of the tangent line to the complex function at a specific point.

2. How do you take the derivative of a complex number?

To take the derivative of a complex number, you need to use the rules of differentiation, which include the power rule, product rule, quotient rule, and chain rule. These rules are applied to the real and imaginary parts of the complex number separately, and then combined to find the derivative of the complex number as a whole.

3. Can you provide an example of finding the derivative of a complex number?

Sure, let's say we have the complex number f(z) = z^2 + 3z + 4. To find its derivative, we first use the power rule to differentiate each term separately: f'(z) = 2z + 3. Then, we combine the real and imaginary parts to get the final derivative of the complex number: f'(z) = 2z + 3i.

4. Are there any special cases when taking the derivative of a complex number?

Yes, there are a few special cases to consider when taking the derivative of a complex number. One is when the complex number has a constant real or imaginary part, in which case the derivative of that part is 0. Another is when the complex number has a constant magnitude, in which case the derivative is equal to the magnitude multiplied by the derivative of the angle of the complex number.

5. How is the derivative of a complex number used in science?

The derivative of a complex number is used in various scientific fields, such as physics, engineering, and economics. It is used to describe the behavior of complex systems and to solve problems involving rates of change, optimization, and control. In particular, it is important in the study of electric circuits, fluid dynamics, quantum mechanics, and financial modeling.

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