Complex Conjugates with sin and cos

In summary, the operator should be applied to the following factor, not the preceding one. The way this is written is a bit confusing to me. An improvement, I believe, would be$$\frac{\partial}{\partial x}(\sqrt{\frac 2 L} \sin(\frac{n\pi x}{L})$$
  • #1
Oaxaca
8
0
I am rather new to the whole idea of complex conjugates and especially operators.

I was trying to understand the solution to a problem I was doing, but the math is confusing me rather than the physics. In the last row of calculations, why does the sin change to a cos, and the d/dx change to what it is. I recognize that n(pi)/L is the first derivative of the inside function for the trig functions, but don't understand how it got there.

Thank you for any help and sorry in advance if I posted this in the wrong section, I'm not positive what section this would fall under.
 

Attachments

  • Screen Shot 2016-02-29 at 8.28.35 PM.png
    Screen Shot 2016-02-29 at 8.28.35 PM.png
    25.1 KB · Views: 2,553
Physics news on Phys.org
  • #2
Oaxaca said:
I am rather new to the whole idea of complex conjugates and especially operators.

I was trying to understand the solution to a problem I was doing, but the math is confusing me rather than the physics. In the last row of calculations, why does the sin change to a cos, and the d/dx change to what it is. I recognize that n(pi)/L is the first derivative of the inside function for the trig functions, but don't understand how it got there.
In the next -to-last line, there is a ##\frac{\partial}{\partial x}## operator. They're taking the partial derivative of what's to the right, with respect to x.
Oaxaca said:
Thank you for any help and sorry in advance if I posted this in the wrong section, I'm not positive what section this would fall under.
 
  • #3
Mark44 said:
In the next -to-last line, there is a ##\frac{\partial}{\partial x}## operator. They're taking the partial derivative of what's to the right, with respect to x.

That makes a lot more sense, I didn't realize that the partial would act as an operator for the proceeding sin. However, how do you choose what to apply the operator to? Is that always the order of the formula?
 
  • #4
Oaxaca said:
That makes a lot more sense, I didn't realize that the partial would act as an operator for the proceeding sin. However, how do you choose what to apply the operator to? Is that always the order of the formula?
The operator should be applied to the following factor, not the preceding one. The way this is written is a bit confusing to me. An improvement, I believe, would be
$$\frac{\partial}{\partial x}(\sqrt{\frac 2 L} \sin(\frac{n\pi x}{L})$$
IOW, without that ) immediately following the partial derivative.
 
  • #5
Mark44 said:
The operator should be applied to the following factor, not the preceding one. The way this is written is a bit confusing to me. An improvement, I believe, would be
$$\frac{\partial}{\partial x}(\sqrt{\frac 2 L} \sin(\frac{n\pi x}{L})$$
IOW, without that ) immediately following the partial derivative.

I got that much (I wrote "proceeding"- improper english on my part, but not a typo :wink:), but I was referring to line six as it seems they move the operator in between the two wave functions, when it was originally outside.
 
Last edited:

Related to Complex Conjugates with sin and cos

What are complex conjugates?

Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. In other words, if a complex number is written as a+bi, its complex conjugate is written as a-bi.

How do complex conjugates relate to the trigonometric functions sin and cos?

When dealing with complex numbers, the trigonometric functions sin and cos can be written in terms of complex numbers as eix = cos(x) + i*sin(x). In this form, the complex conjugate of eix is e-ix, which corresponds to cos(x) - i*sin(x).

Why are complex conjugates important in mathematics?

Complex conjugates are important because they allow us to simplify complex numbers and make calculations easier. They also have many applications in fields such as engineering, physics, and signal processing.

How do we find the complex conjugate of a given complex number?

To find the complex conjugate of a given complex number, we simply change the sign of its imaginary part. For example, the complex conjugate of 3+4i is 3-4i.

Are complex conjugates always in the form of a+bi?

No, complex conjugates can also be in the form of a-bi, depending on the original complex number. It is important to remember that the real part remains the same, but the imaginary part changes in sign when taking the complex conjugate.

Similar threads

Replies
4
Views
450
Replies
3
Views
1K
  • Electrical Engineering
Replies
5
Views
623
  • Introductory Physics Homework Help
Replies
3
Views
246
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
855
Replies
3
Views
1K
  • Differential Geometry
Replies
1
Views
1K
Replies
4
Views
2K
Back
Top