Is my proof of the identity $|e^{iy}|=1$ logically valid?

Question

Prove: $|e^{iy}|=1$ for all real $y$.

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$$|z|=zz^*$$ $$|e^{iy}|=e^{iy}e^{-iy}=e^0=1$$

To me, this seems fine but my concern is that I define $z$ as the entire value inside the absolute value brackets, yet the only complex number in line 2 is $iy$. Is this correct?

Answer 1

If you've not yet shown that the conjugate of $e^{iy}$ is equal to $e^{-iy}$ (as fleablood encouraged), then perhaps recall that \begin{equation} e^{iy}=\cos(y) + i\sin(y) \end{equation} and compute the modulus of $e^{iy}$ from here.

Answer 2

You've got a mistake right at the start - $|z|^2 = z\bar{z}$. However, once you fix that then all you need to do is state that $|z| \geq 0$ so that when you take the square root you don't have to worry about the negative solution. Also, you need to have proven that $\overline{e^{iy}} = e^{-iy}$, because it doesn't flow directly from the definition of the complex conjugate.

Answer 3

I feel very uneasy with this because it isn't clear to what degree we are expected to have internalized the polar notation of complex numbers. Given $z = e^{x + iy}$ it's darn near a fundamental given that $|z| = e^x$. So ... what's the issue?

But if we are new to polar notation and need to remind ourselves that $z = e^{x+iy} = e^x(\cos y + i\sin y)$ then proving that $|z| = e^x|\cos y + i \sin y| = e^x$ can be done any number of ways. Including noting $\overline {e^{x+iy}} = e^x\overline{\cos x + i \sin y} = e^x(\cos x - i\sin x) = e^x(\cos x + i \sin(-x)) = e^{x - iy}$. [So $e^{x + iy}\overline {e^{x+iy}} = (e^x)^2$ so $|z| = \sqrt {z\overline{z}} = e^x$ ]

[Assuming you have proven $e^z e^w = e^{z + w}$ for complex $w, z$... which is something else that may be proven.]

[In all, I recommend proving every fundamental fact of polar notation at once and getting it over with. It depends on your analysis course and text.]

Answer 4

1. First way, when $\text{z}\in\mathbb{C}$: $$\left|e^{\text{z}i}\right|=\left|e^{\left(\Re[\text{z}]+\Im[\text{z}]i\right)i}\right|=\left|e^{-\Im[\text{z}]+\Re[\text{z}]i}\right|=\left|e^{-\Im[\text{z}]}\right|\cdot\left|e^{\Re[\text{z}]i}\right|=\left|e^{-\Im[\text{z}]}\right|=\frac{1}{e^{\Im[\text{z}]}}$$

Because with Euler's formula: $\left|e^{\Re[\text{z}]i}\right|=\sqrt{\cos^2(\Re[\text{z}])+\sin^2(\Re[\text{z}])}=1$.

When $\Im[\text{z}]=0$ we get:

$$\left|e^{\text{z}i}\right|=\frac{1}{e^0}=\frac{1}{1}=1$$

2. Second way, when $\text{z}\in\mathbb{C}$: $$\text{z}\overline{\text{z}}=|\text{z}|^2\to e^{\text{z}i}\cdot\overline{e^{\text{z}i}}=\left(e^{-\Im[\text{z}]}\right)\cdot\left(e^{\Re[\text{z}]i}\right)\cdot\overline{\left(e^{-\Im[\text{z}]}\right)\cdot\left(e^{\Re[\text{z}]i}\right)}=$$ $$\left(e^{-\Im[\text{z}]}\right)\cdot\left(e^{\Re[\text{z}]i}\right)\cdot\left(e^{-\Im[\text{z}]}\right)\cdot\left(e^{-\Re[\text{z}]i}\right)=e^{-2\Im[\text{z}]}\cdot e^{\Re[\text{z}]i-\Re[\text{z}]i}=e^{-2\Im[\text{z}]}\cdot e^{0}=e^{-2\Im[\text{z}]}$$

When $\Im[\text{z}]=0$ we get:

$$e^{-2\cdot0}=e^0=1$$