# How to make the conjugate transpose

I have a problem to make the conjugate and transpose the matrix.May anyone help me please.

MatrixForm[{{E^(I*β1 + I*β3) Cos[β2],
E^(I β1 -
I*β3) Sin[β2]}, {(-E^((-I) β1 +
I*β3)) Sin[β2],
E^((-I) β1 - I*β3)*Cos[β2]}}]


and I get the output

{{E^((-I)*Conjugate[β1] - I*Conjugate[β3])*
Conjugate[Cos[β2]],
-(E^(I*Conjugate[β1] - I*Conjugate[β3])*
Conjugate[Sin[β2]])},
{E^((-I)*Conjugate[β1] + I*Conjugate[β3])*
Conjugate[Sin[β2]],
E^(I*Conjugate[β1] + I*Conjugate[β3])*
Conjugate[Cos[β2]]}}


my problem here, I just want to conjugate the imaginary part but it conjugate all include the angle Cos and Sin (real). How can I solve this.

Thank you

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## 1 Answer

m = {{E^(I*β1 + I*β3) Cos[β2],
E^(I β1 - I*β3) Sin[β2]}, {(-E^((-I) β1 + I*β3)) Sin[β2],
E^((-I) β1 - I*β3)*Cos[β2]}};

MatrixForm[
Assuming[{β1, β2, β3} ∈ Reals,
Simplify@ConjugateTranspose[m]]]


$$\left( \begin{array}{cc} e^{i \text{\beta 1}+i \text{\beta 3}} \cos (\text{\beta 2}) & e^{i \text{\beta 1}-i \text{\beta 3}} \sin (\text{\beta 2}) \\ -e^{i \text{\beta 3}-i \text{\beta 1}} \sin (\text{\beta 2}) & e^{-i \text{\beta 1}-i \text{\beta 3}} \cos (\text{\beta 2}) \\ \end{array} \right)$$