Algebraic Equation?

Question

$$Ve^{i\theta} = We^{i\phi}$$

where, $V$ and $W$ are some real constants. From this my book concludes: $\theta = \phi$. How does it conclude this? I don't see why its valid to just equate the exponent parts (which I think my book does).

Answer 1

If $R>0$ and $\theta\in \mathbb{R}$ then the complex number $Re^{i\theta}$ has magnitude $R$ and angle $\theta$. So if $V>0$ and $W>0$ and $Ve^{i\theta}=We^{i\phi}$ then the magnitudes and angles are equal. Hence $V = W$ and $\theta = \phi +2\pi k$ for some $k \in \mathbb{Z}$.

Answer 2

The conclusion is false anyways, but I'll show you how to deduce something close to what you really need.

$$Ve^{i\theta}=We^{i\phi}$$ $$|Ve^{i\theta}|=|We^{i\phi}|$$ $$|V||e^{i\theta}|=|W||e^{i\phi}|$$

Since $|e^{i\gamma}|=1$ for every $\gamma$, therefore:

$$|V|=|W|$$

Case 1: $V,W>0$ I think this is the case in your book, but they don't stress the importance that this is the case for their argument because their argument is sloppy.

In this case $V=|V|=|W|=W$, therefore you can divide in your original equation by $V$. To get $e^{i\theta}=e^{i\phi}$. For circuit analysis purposes, all you would need to know is this. Which is equivalent to $\frac{\theta-\phi}{2\pi}$ is an integer (and this is all what you really need to know when working with impedances). but NOT $\theta=\phi$.

Case 2: $V,W=0$

Then there is no relation between $\theta,\pi$

Case 3: $V,W$ are non-zero reals

Then one can conclude that $\frac{\theta-\phi}{\pi}$ is an integer by the method I used earlier. I'll leave this to you

Answer 3

You have that $$ e^{i\alpha} = \cos \alpha + i\sin\alpha \quad\text{and}\quad |e^{i\alpha}| = \sqrt{(\cos\alpha)^2 + (\sin\alpha)^2} = 1\text{,} $$

i.e. $e^{i\alpha}$ always is a point on the unit circle, at angle $\alpha$ away counter-clockwise from $+1$.

If $V,W$ are both positive real numbers, then the only way for $$ Ve^{i\theta} = We^{i\phi} $$ to hold is thus that $$ V = W \quad\text{ and }\quad e^{i\theta} = e^{i\phi} \text{.} $$ That does not implies that $\theta = \phi$, though. $\alpha \to e^{i\alpha}$ is not an injective function, because $$ e^{i\alpha} = e^{i\alpha + i2\pi n} \quad\text{for $n \in \mathbb{Z}$.} $$ But what you can say is that $Ve^{i\theta} = We^{i\phi}$ implies $$ V = W \quad\text{ and }\quad \theta - \phi = 2\pi n \text{ for some $n \in \mathbb{Z}$.} $$ In your book, $\theta$ and $\phi$ are probably supposed to be angles, and thus are either assumed lie within the interval $[0,2\pi)$ or it is implicitly assumed that two angles are equal if their difference is a multiple of $2\pi$. With either of this additional prerequisits, you can indeed conclude that $\theta = \phi$.