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I'm having a little hard time understanding the equation:

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Why Euler's relation's square is equal to 1?

I know how to write Euler's equation in terms of sin and cos:

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But when I do the math, the square of this equality is nothing but 1. I can't get rid of from the sin and cos in the equation: $$(e^{jwt})^2 = cos^2(jwt)+2jcos(jwt)sin(jwt)+j^2sin^2(jwt)$$

$$= cos^2(jwt)-sin^2(jwt)+2jcos(jwt)sin(jwt)$$

I'm probably missing something with the Euler's relation here so I would like to hear what am I missing.

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2 Answers 2

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Notice, that \$\forall x\in\mathbb{R}\$:

$$\left|\exp\left(xi\right)\right|=1\tag1$$

Because:

$$\left|\exp\left(xi\right)\right|=\left|\cos\left(x\right)+\sin\left(x\right)i\right|=\sqrt{\underbrace{\cos^2\left(x\right)+\sin^2\left(x\right)}_{=\space 1}}=\sqrt{1}=1\tag2$$

So, for your integral we get:

\begin{equation} \begin{split} \mathcal{I}\left(\text{T}_0\right)&=\int\limits_0^{\text{T}_0}\left|\exp\left(\omega_0t\text{j}\right)\right|^2\space\text{d}t\\ \\ &=\int\limits_0^{\text{T}_0}\left|\cos\left(\omega_0t\right)+\sin\left(\omega_0t\right)\text{j}\right|^2\space\text{d}t\\ \\ &=\int\limits_0^{\text{T}_0}\left(\sqrt{\underbrace{\cos^2\left(\omega_0t\right)+\sin^2\left(\omega_0t\right)}_{=\space1}}\right)^2\space\text{d}t\\ \\ &=\int\limits_0^{\text{T}_0}\underbrace{\left(\sqrt{1}\right)^2}_{=\space1}\space\text{d}t\\ \\ &=\int\limits_0^{\text{T}_0}1\space\text{d}t\\ \\ &=\left[t\right]_0^{\text{T}_0}\\ \\ &=\text{T}_0-0\\ \\ &=\text{T}_0 \end{split}\tag3 \end{equation}

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I believe the vertical bars inside the integration denote magnitude. This exponent is a phasor whose magnitude is one and angle is (wt). The square of one is also one.

The magnitude of the phasor (or complex number) is the square root of: the real part squared plus the imaginary part (without j) squared. When you add sin^2 and cos^2, the result is one.

Hope this helps.

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    \$\begingroup\$ Yes yes, I see what I was missing, I didn't pay attention to the absolute value inside the integration, now it makes sense. I was focused on the square part. \$\endgroup\$
    – Berk
    Feb 20, 2021 at 15:13

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