Tan (a + ib) = x + iy, so how is tan (a-ib) = x - iy

  • Thread starter jaus tail
  • Start date
  • Tags
    Ib Tan
In summary, Tan(x+iy) can be written in terms of a REAL power series and the complex conjugate of tan(x+iy) is tan(x-iy).
  • #1
jaus tail
615
48

Homework Statement


upload_2018-2-2_10-34-31.png


Homework Equations


I didn't understand the first step.
If tan (z) = a + ib, how can tan (z conjugate) be a - ib?
tan is not a linear function.
I know conjugates. x + iy, conjugate is x - iy
But here the tan function is there.

The Attempt at a Solution


...
 

Attachments

  • upload_2018-2-2_10-34-31.png
    upload_2018-2-2_10-34-31.png
    2.7 KB · Views: 2,459
  • Like
Likes Somali_Physicist
Physics news on Phys.org
  • #2
jaus tail said:

Homework Statement


View attachment 219526

Homework Equations


I didn't understand the first step.
If tan (z) = a + ib, how can tan (z conjugate) be a - ib?
tan is not a linear function.
I know conjugates. x + iy, conjugate is x - iy
But here the tan function is there.

The Attempt at a Solution


...
I suspect, but haven't confirmed this, that they are converting ##\tan(x + iy)## to ##\frac{\sin(x + iy)}{\cos(x + iy)}##, and then expanding both numerator and denominator using the addition formulas for sine and cosine. You'll have terms that involve sin(iy) and cos(iy). These can be rewritten in terms of exponentials.

We know that ##e^{ix} = \cos(x) + i\sin(x)## and that ##e^{-ix} = \cos(x) - i\sin(x)##. These can be solved for cos(x) and sin(x). To get cos(iy) and sin(iy), just substitute iy in place of x in the exponential equations for cos(x) and sin(x).

That would be my approach.
 
  • Like
Likes jaus tail
  • #3
I guess it's for general function then.
Like if f(x + iy) = real + imaginary times i
then f(x - iy) = real - imaginary times i
right?
 
  • #4
jaus tail said:
I guess it's for general function then.
Like if f(x + iy) = real + imaginary times i
then f(x - iy) = real - imaginary times i
right?

That is only true if the power series expansion of ##f## contains only real coefficients. So in the case of ##f(z)=\tan(z)##, it works. In a case like ##f(z)=iz##, it does not.
 
  • Like
Likes PeroK and jaus tail
  • #5
i think I have found a solution, has OP found one?Or was his function statement solution.
 
  • Like
Likes jaus tail
  • #7
jaus tail said:
Thanks. It's a solved example. Have you also used the method in this link
http://mathumatiks.org/subpage-556-Real and Imaginary Parts of Circular .htm
Mathumatiks Website.

If the question you are asking is why ##\tan(x+iy)=a+bi## implies that ##\tan(x-iy)=a-bi##, the answer is to look at the nature of the taylor series of ##\tan## As I said before. If that's not your question, then what is?
 
Last edited:
  • #8
Dick said:
If the question you are asking is why ##\tan(x+iy)=x+bi## implies that ##\tan(x-iy)=x-bi##, the answer is to look at the nature of the taylor series of ##\tan## As I said before. If that's not your question, then what is?
I do not know how to prove for any function f(x+iy) that
complex conjugate of f(x+iy) = f(x-iy). But one can easily prove that complex conjugate of sum, difference, product and division of two complex numbers will be the sum, difference, product and division respectively of the complex conjugate. Now if series expansion for tan(x+iy) is valid then it follows that complex conjugate of tan(x+iy) = tan(x-iy).
 
  • Like
Likes jaus tail
  • #9
I didn't understand how
Tan(X + iy) can be a + ib
N then
Tan(x-iy) be a-ib
I didn't know tan is linear function.
 
  • #10
Let'sthink said:
I do not know how to prove for any function f(x+iy) that
complex conjugate of f(x+iy) = f(x-iy). But one can easily prove that complex conjugate of sum, difference, product and division of two complex numbers will be the sum, difference, product and division respectively of the complex conjugate. Now if series expansion for tan(x+iy) is valid then it follows that complex conjugate of tan(x+iy) = tan(x-iy).
Yes I was struggling with this.
 
  • #11
jaus tail said:
I didn't understand how
Tan(X + iy) can be a + ib
N then
Tan(x-iy) be a-ib
I didn't know tan is linear function.
I am not saying that it is linear function. But if it can be written in terms of power series then it follows that complex conjugate of tan(x+iy) = tan (x-iy).
 
  • Like
Likes jaus tail
  • #12
Let'sthink said:
I am not saying that it is linear function. But if it can be written in terms of power series then it follows that complex conjugate of tan(x+iy) = tan (x-iy).

I hope you mean "if it can be written in terms of a REAL power series".
 
  • Like
Likes jaus tail
  • #13
Dick said:
I hope you mean "if it can be written in terms of a REAL power series".
No no complex number series with real coefficients. for example sin theta = theta - theta^3/3! etc is valid for complex number z too as
sin z = z - z^3/3! etc
 
  • Like
Likes jaus tail
  • #14
jaus tail said:
Yes I was struggling with this.
The suggestion of Mark44 in post #2 is probably the easiest way to go. Have you tried it?
 
  • Like
Likes jaus tail
  • #15
Let'sthink said:
No no complex number series with real coefficients. for example sin theta = theta - theta^3/3! etc is valid for complex number z too as
sin z = z - z^3/3! etc

That is what I meant. See post #4. I'm a little vague about what the actual question here is. The reference is a solved example. What part isn't clear?
 
  • Like
Likes jaus tail
  • #16
jaus tail said:
I didn't understand how
Tan(X + iy) can be a + ib
N then
Tan(x-iy) be a-ib
I didn't know tan is linear function.
It's not but you can show that through series that it works out.Or do the gruelling separation of tan(x) and rearranging to get tan(a-ib)
 
  • Like
Likes jaus tail
  • #17
Since I helped OP with an algebraic combinatorics problem not too long ago, I thought a straightforward matrix approach may be enlightening here. (And since we're considering a solved example I don't think I'm overstepping forum rules)

- - - -

use the fact that any complex number can be represented by a 2 x 2 matrix with real entries. This gives us:
##\mathbf Z = \begin{bmatrix}
x & -y\\
y & x
\end{bmatrix}= x
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix} +
y \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
= x\mathbf I + y\mathbf i##

note that ##\mathbf Z## commutes with any other complex number represented as a 2 x 2 matrix. This allows us to treat ##\mathbf Z## as a 'regular number' (and immediately gives rise to ##\mathbf Z## being normal).

Then use the fact that tangent function can be written as ##\tan(u) = \sin(u)\big(\cos(u)\big)^{-1}## for some real ##u##, and both functions have entirely real power series so we know that tangent can be written entirely in reals. So we restrict ourselves to the field of reals -- specifically real 2 x 2 matrices.

##
\tan\big( x\mathbf I + y\mathbf i\big) = \tan\big(\mathbf Z\big) = \begin{bmatrix}
a & -b\\
b & a
\end{bmatrix} = a \mathbf I + b\mathbf i = \sin(\mathbf Z)\big(\cos(\mathbf Z)\big)^{-1} = \big(\cos(\mathbf Z)\big)^{-1}\sin(\mathbf Z) ##

(Justification for commuting at the end is simply that ##\mathbf Z## can be treated like a 'regular number'. This gets you 99% there. If more justification is needed, do a relaxation and allow the scalar field to have complex numbers-- ##\mathbf Z## being normal must be diagonalizable and the result easily follows since both matrices may be chosen to have the same unitary basis of eigenvectors and diagonal matrices commute.)

use of the transpose gives

##
\tan\big(\mathbf Z\big)^T = \begin{bmatrix}
a & -b\\
b & a
\end{bmatrix}^T =\begin{bmatrix}
a & b\\
-b & a
\end{bmatrix} = a \mathbf I - b\mathbf i ##

##\tan\big(\mathbf Z\big)^T = \Big(\sin(\mathbf Z)\big(\cos(\mathbf Z)\big)^{-1}\Big)^T = \big(\cos(\mathbf Z)^T\big)^{-1} \big(\sin(\mathbf Z)\big)^T =\big(\cos(\mathbf Z^T)\big)^{-1} \big(\sin(\mathbf Z^T)\big) = \tan\big(\mathbf Z^T\big) = \tan\big(x\mathbf I - y\mathbf i\big)##

thus
##a \mathbf I - b\mathbf i = \tan\big(x\mathbf I - y\mathbf i\big)##
 
  • Like
Likes jaus tail
  • #18
Ray Vickson said:
The suggestion of Mark44 in post #2 is probably the easiest way to go. Have you tried it?
Yeah that's actually pretty easy. I went ahead with that method and got the answer. Thanks.
 
  • #19
StoneTemplePython said:
Since I helped OP with an algebraic combinatorics problem not too long ago, I thought a straightforward matrix approach may be enlightening here. (And since we're considering a solved example I don't think I'm overstepping forum rules)

- - - -

use the fact that any complex number can be represented by a 2 x 2 matrix with real entries. This gives us:
##\mathbf Z = \begin{bmatrix}
x & -y\\
y & x
\end{bmatrix}= x
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix} +
y \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
= x\mathbf I + y\mathbf i##

note that ##\mathbf Z## commutes with any other complex number represented as a 2 x 2 matrix. This allows us to treat ##\mathbf Z## as a 'regular number' (and immediately gives rise to ##\mathbf Z## being normal).

Then use the fact that tangent function can be written as ##\tan(u) = \sin(u)\big(\cos(u)\big)^{-1}## for some real ##u##, and both functions have entirely real power series so we know that tangent can be written entirely in reals. So we restrict ourselves to the field of reals -- specifically real 2 x 2 matrices.

##
\tan\big( x\mathbf I + y\mathbf i\big) = \tan\big(\mathbf Z\big) = \begin{bmatrix}
a & -b\\
b & a
\end{bmatrix} = a \mathbf I + b\mathbf i = \sin(\mathbf Z)\big(\cos(\mathbf Z)\big)^{-1} = \big(\cos(\mathbf Z)\big)^{-1}\sin(\mathbf Z) ##

(Justification for commuting at the end is simply that ##\mathbf Z## can be treated like a 'regular number'. This gets you 99% there. If more justification is needed, do a relaxation and allow the scalar field to have complex numbers-- ##\mathbf Z## being normal must be diagonalizable and the result easily follows since both matrices may be chosen to have the same unitary basis of eigenvectors and diagonal matrices commute.)

use of the transpose gives

##
\tan\big(\mathbf Z\big)^T = \begin{bmatrix}
a & -b\\
b & a
\end{bmatrix}^T =\begin{bmatrix}
a & b\\
-b & a
\end{bmatrix} = a \mathbf I - b\mathbf i ##

##\tan\big(\mathbf Z\big)^T = \Big(\sin(\mathbf Z)\big(\cos(\mathbf Z)\big)^{-1}\Big)^T = \big(\cos(\mathbf Z)^T\big)^{-1} \big(\sin(\mathbf Z)\big)^T =\big(\cos(\mathbf Z^T)\big)^{-1} \big(\sin(\mathbf Z^T)\big) = \tan\big(\mathbf Z^T\big) = \tan\big(x\mathbf I - y\mathbf i\big)##

thus
##a \mathbf I - b\mathbf i = \tan\big(x\mathbf I - y\mathbf i\big)##
Love this for it's complex simplicity

:sorry:
 
  • Like
Likes Let'sthink
  • #20
Somali_Physicist said:
Love this for it's complex simplicity

:sorry:

The approach, though not the problem, comes from Stillwell's Naive Lie Theory. It turns out that sometimes representing complex numbers (and also quaternions) as 2 x 2 matrices can be quite useful. In general solving a problem in multiple ways helps people make insights and connections. I didn't think OP would see this approach elsewhere, that's why I included it.
 
  • Like
Likes Somali_Physicist

1. What does the equation "Tan (a + ib) = x + iy" mean?

The equation represents the tangent function of a complex number, where a + ib is the real and imaginary components of the complex number and x + iy is the resulting complex number after applying the tangent function.

2. How is the tangent function applied to complex numbers?

The tangent function can be applied to complex numbers by converting the complex number into its polar form, where the magnitude represents the distance from the origin and the argument represents the angle. The tangent function can then be applied to the argument to obtain the resulting complex number.

3. What is the difference between "Tan (a + ib)" and "Tan (a-ib)"?

The only difference between the two equations is the sign of the imaginary component. In the first equation, the imaginary component is positive (ib), while in the second equation, it is negative (-ib). This difference only affects the resulting complex number, not the application of the tangent function.

4. Why is the equation "Tan (a-ib) = x - iy" useful?

This equation is useful in solving complex trigonometric equations and in representing the tangent function on the complex plane. It also has applications in fields such as physics, engineering, and mathematics.

5. Are there any other trigonometric functions that can be applied to complex numbers?

Yes, the sine, cosine, and inverse trigonometric functions can also be applied to complex numbers. They follow the same principles as the tangent function, where the complex number is converted to polar form and the function is applied to the argument.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
857
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
882
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top