Complex Conjugate in the form $x+iy$

Question

I have this complex number: $$e^{t \sin \phi - i \tan \phi} $$

I want to find its conjugate however I am confused by the form it is currently in. I was thinking of it to be similar to $re^{i\theta}$ , with $r$ being $1$ here, but the $t\sin$ in the exponent is giving me issues. In addition, how would I get the trig functions into $x+iy$? Normally I would have done $re^{i\theta}$ which would then be $r(\cos \theta + i \sin \theta)$.

Answer 1

$$\overline{e^z}=e^{\overline z}$$

because

$$\overline{e^z}=e^{x}\overline{(\cos y+i\sin y)}=e^{x}{(\cos\ y-i\sin y)}=e^{\overline z}.$$

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More generally, if $f$ is an holomorphic function taking real values on the real axis, $\overline{f(z)}=f(\overline z)$.

Answer 2

So you've got $$e^{t \sin \phi - i \tan \phi} = \underbrace{e^{-i \tan \phi}}_{\text{complex}}\underbrace{e^{t \sin \phi}}_{\text{real}}$$ So the complex conjugate here would be \begin{equation} e^{i \tan \phi} e^{t \sin \phi} \end{equation} i.e. where we have used $y^* = \alpha^* z^* = \alpha z^*$ since $\alpha$ is real. and the conjugate of $z = e^{i \tan \phi}$ is $z^* = e^{-i \tan \phi}$

Answer 3

We have that

$$e^{t\sin \phi-i\tan \phi}=e^{t\sin \phi}\cdot e^{-i\tan \phi}$$

and therefore

$$\overline{e^{t\sin \phi-i\tan \phi}}=\overline{e^{t\sin \phi}\cdot e^{-i\tan \phi}}=e^{t\sin \phi}\cdot \overline{e^{-i\tan \phi}}=e^{t\sin \phi}\cdot e^{i\tan \phi}=e^{t\sin \phi+i\tan \phi}$$

indeed

$$e^{i\theta}=\cos \theta+i\sin \theta \implies \overline{e^{i\theta}}=\overline{\cos \theta+i\sin \theta}=\cos \theta-i\sin \theta=e^{-i\theta}$$