# Complex numbers and $e^{i\theta}$

I have a complex $$z$$ that I want to write as $$z=|z|e^{i\theta}$$ where $$\theta$$ is its phase.

I tried using 2 "different" ways and got different results, so I assume one of them is wrong.

This is the number: $$z=\frac{a}{b-ic}$$

where $$a,b,c \in \mathbb{R}$$ and positive.

Now we can write:

$$z=\frac{a}{b-ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(0-\arctan\left(\frac{-c}{b}\right)\right)}$$

But also:

$$z=\frac{-a}{-b+ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(\pi-\arctan\frac{c}{-b}\right)}$$

So where is the mistake?

• @vitamind Sorry, done. Feb 20, 2021 at 18:27
• Don't you think rewriting the expression would make it a lot easier to understand and solve the problem? Your number breaks down to $$\frac{a}{b-ic},\quad a,b,c\in\mathbb{R}.$$ Feb 20, 2021 at 18:33
• @vitamind Well, I guess you're right. I'll edit Feb 20, 2021 at 18:37
• In any case you need to have a discussion about the sign of $w^2LC-1$, and determine quadrant the phase belongs to, or use $\operatorname{atan2}$ function instead of $\arctan$.
– zwim
Feb 20, 2021 at 18:38

I think this was actually a better question when it still had all the variables $$\omega$$, $$R$$, $$L$$, and $$C.$$ Eliminating those variables eliminates one of the sources of ambiguity inherent in the original problem, which is that we don't know which of the two quantities $$\omega^2 RLC$$ or $$R$$ is greater.

The problem is the in the conversion of $$p + iq$$ to $$\sqrt{p^2+q^2} \, e^{i \arctan(q/p)}.$$ That formula is correct only when $$p > 0.$$ If $$p < 0$$ a correct formula is $$p + iq = \sqrt{p^2+q^2} \, e^{i (\pi + \arctan(q/p))}.$$

And of course if $$p = 0$$ neither formula works at all.

So if you have $$b - ic,$$ where $$b$$ and $$c$$ are positive, you have the case $$p = b > 0$$, so $$b - ic = \sqrt{b^2+c^2} \, e^{i \arctan(-c/b)}.$$

But when you look at $$-b + ic,$$ you now have the case $$p = -b < 0,$$ so $$-b + ic = \sqrt{b^2+c^2} \, e^{i (\pi + \arctan(-c/b))}.$$

You applied the wrong formula to the case $$-b + ic,$$ and that's where the extra $$\pi$$ comes from.

In your original exercise, you had the expression $$\left(R - \omega^{2}RLC\right)+i\omega L$$ in one case and $$\left(\omega^{2}RLC - R\right)-i\omega L$$ in the other. If $$R - \omega^{2}RLC > 0,$$ then $$\omega^{2}RLC - R < 0,$$ and vice versa. That is, one of these expressions is the $$p > 0$$ case and the other is the $$p < 0$$ case. The plain arc tangent formula only works in one case; the other case needs some modification, such as adding $$\pi$$ to the arc tangent. (Subtracting $$\pi$$ works equally well, because that's the same as adding $$\pi$$ and then subtracting $$2\pi$$.)

That's why there's a difference of $$\pi$$ between the angles you found in your results. The correct result is the one where you applied your formula to $$p + iq$$ where $$p > 0.$$ In the other case your angle is shifted by $$\pi$$ radians.

• I see. So in the last example that I gave it should be $z=\frac{-a}{-b+ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(\pi-\left(\pi+\arctan\left(\frac{c}{-b}\right)\right)\right)}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(-\arctan\left(-\frac{c}{b}\right)\right)}$. In your example, as you said, for $a+ib$ in order to use the formula without $\pi$ we need $a$ to be positive. Are there any conditions for $b$ ? we dont care about the sign of $b$ ? Thanks. Feb 20, 2021 at 19:00
• Note that I wrote the answer with one set of variables and then edited it to use a different set of variables to avoid names that contradicted the edited question. In the updated notation, converting $p + iq$ to the form $re^{i\theta},$ only the sign of $p$ matters. The value of $q$ can be anything: positive, negative, or zero. The key fact is that $\arctan$ always gives an angle pointing into the right half of the complex plane, but has no difficulty correctly identifying whether it should be above, below, or on the real axis. Feb 20, 2021 at 19:07
• The only problem really is when we need the angle to point along the imaginary axis or into the left half of the plane. Then $\arctan$ gives the wrong direction, but since $(-q)/(-p) = q/p$ it's simply exactly in the opposite direction we want, so adding or subtracting $\pi$ will fix it. Feb 20, 2021 at 19:13

Somewhat off topic:

Since the OP's question was answered by David K's response, I consider it open season to discuss a more straightforward approach.

Without loss of generality, $$(b - ic) \neq [0 - i(0)] \implies (b^2 + c^2) \neq 0.$$

$$z = \frac{a}{b-ic} \times \frac{b+ic}{b+ic} = \frac{a(b + ic)}{b^2 + c^2}.$$

Since $$(b^2 + c^2) \in \mathbb{R^+}$$, you can let $$s\in\mathbb{R}$$ represent $$\frac{a}{b^2 + c^2}$$, and then let $$z_0$$ represent $$\left(\frac{1}{s} \times z\right).$$ Note that since $$(a)$$ is presumed to be positive, $$s$$ must be positive.

Therefore, the problem reduces to expressing

$$(b + ic) = z_0 = re^{i\theta}.$$

Once this is done, you will have that $$z = (sr)e^{i\theta}.$$

Let $$r$$ denote $$\sqrt{b^2 + c^2}$$ and
let $$\theta$$ denote the unique angle, within a modulus of $$(2\pi)$$
such that $$\cos(\theta) = \frac{b}{r}$$ and $$\sin(\theta) = \frac{c}{r}.$$

Then $$z_0 = (b + ic) = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$$