0
$\begingroup$

I am trying to proof that if $e^{i\bar{z}}=e^{iz}$ then $x=n\pi$.

$$e^{i\bar{z}}=e^{iz}$$ $$e^{i(x-iy)}=e^{i(x+iy)}$$ $$e^{ix}e^{y}=e^{ix}e^{-y}$$

then $e^{ix}=e^{ix}$ and $e^{y}=e^{-y}$.

this occurs if $y=0$ and $x=x+2\pi n$.

I am not sure how to continue to show that $x=n\pi$.

  • 1
    I'm not sure the problem is stated correctly. Take $z=1$ (or any real number). Then $e^i=e^i$. – 2017-01-17
  • 2
    As you've stated the problem, it's only necessary that $z$ be real, i.e. $y=0.$ $e^{ix}= e^{ix}$ always... no extra conditions needed. Sure you don't want to conjugate the whole thing, or something? – 2017-01-17
  • 0
    @ElliotG, that is not what the question asks. I am required to show that $x=n\pi$ where $n$ is a positive integer. – 2017-01-17
  • 0
    @ElliotG Then the question is wrong. You cannot show that, because it is false. – 2017-01-17
  • 0
    Perhaps you mean $e^{iz}=e^{\overline{iz}}$? – 2017-01-17
  • 0
    @StellaBiderman, this probably correct. My textbook font is not clear. – 2017-01-17
  • 0
    Possible duplicate of [Proof that $e^{i\bar{z}}=\overline{e^{iz}}$ if and only if $z=k\pi\in Z$](http://math.stackexchange.com/questions/646845/proof-that-ei-barz-overlineeiz-if-and-only-if-z-k-pi-in-z) – 2017-01-17
  • 0
    @Isaac in that case I'm voting to close as a duplicate, as that question has been answered here: http://math.stackexchange.com/questions/646845/proof-that-ei-barz-overlineeiz-if-and-only-if-z-k-pi-in-z?rq=1 – 2017-01-17

1 Answers 1

0

Well, when $\exists\space\text{z}\in\mathbb{C}$:

  1. $$\exp\left(\text{z}i\right)=\exp\left[\Re\left[\text{z}\right]i\right]\cdot\exp\left[-\Im\left[\text{z}\right]\right]=\frac{\cos\left(\Re\left[\text{z}\right]\right)+\sin\left(\Re\left[\text{z}\right]\right)i}{\exp\left[\Im\left[\text{z}\right]\right]}\tag1$$
  2. $$\exp\left(\overline{\text{z}}i\right)=\exp\left[\Re\left[\text{z}\right]i\right]\cdot\exp\left[\Im\left[\text{z}\right]\right]=\frac{\cos\left(\Re\left[\text{z}\right]\right)+\sin\left(\Re\left[\text{z}\right]\right)i}{\exp\left[-\Im\left[\text{z}\right]\right]}\tag2$$

So, when:

$$\exp\left(\text{z}i\right)=\exp\left(\overline{\text{z}}i\right)\space\Longleftrightarrow\space\frac{\cos\left(\Re\left[\text{z}\right]\right)+\sin\left(\Re\left[\text{z}\right]\right)i}{\exp\left[\Im\left[\text{z}\right]\right]}=\frac{\cos\left(\Re\left[\text{z}\right]\right)+\sin\left(\Re\left[\text{z}\right]\right)i}{\exp\left[-\Im\left[\text{z}\right]\right]}\tag3$$

Which leads towards:

$$\exp\left[\Im\left[\text{z}\right]\right]=\exp\left[-\Im\left[\text{z}\right]\right]\tag4$$