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

[jaus tail]

Homework Statement

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

...

 

[Mark44]

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.

 

[jaus tail]

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?

 

[Dick]

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.

 

[Somali_Physicist]

i think I have found a solution, has OP found one?Or was his function statement solution.

 

[jaus tail]

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.

 

[Dick]

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?

 

[Let'sthink]

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).

 

[jaus tail]

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.

 

[jaus tail]

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.

 

[Let'sthink]

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).

 

[Dick]

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".

 

[Let'sthink]

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

 

[Ray Vickson]

Yes I was struggling with this.

The suggestion of Mark44 in post #2 is probably the easiest way to go. Have you tried it?

 

[Dick]

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?

 

[Somali_Physicist]

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)

 

[StoneTemplePython]

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
$a \mathbf I - b\mathbf i = \tan\big(x\mathbf I - y\mathbf i\big)$

 

[jaus tail]

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.

 

[Somali_Physicist]

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

 

[StoneTemplePython]

Love this for it's complex simplicity

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.