# Converting rectangular form complex numbers to polar form in different quadrants

I know how to convert a complex number from rectangular form to polar form, but I can't understand when should I add or subtract $$\pi$$ from/to arctan in different quadrants.

Can some one explain this for all quadrants? (maybe except for I)

• Before adding or subtracting you should decide what range of the angle is your choice. Generally two variants are in common use: $[0,2\pi)$ and $(-\pi,\pi]$.
– user
Mar 24, 2021 at 21:52

Normally, I'm not supposed to answer, because the OP hasn't shown any work. But this is a different situation, so if I get downvoted, okay. The way that I was taught ("An Introduction To Complex Function Theory" : Bruce Palka) you are supposed to use the sine and cosine functions.

Let $$z = (x + iy)$$, where $$x$$ and $$y$$ are not both $$= 0$$.

Let $$r = +\sqrt{x^2 + y^2}$$.

Then, within a modulus of $$(2\pi)$$, there will be a unique angle $$\theta$$, such that

$$\cos(\theta) = \frac{x}{r}$$ and $$\sin(\theta) = \frac{y}{r}.$$

Then, $$(x + iy) = z = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$$

Therefore, you never have to concern yourself with the tangent function, or wondering which quadrant $$\theta$$ is in. The quadrant is determined by whether (for example) $$\frac{x}{r}$$ and/or $$\frac{y}{r}$$ is positive.

Responding to the OP's comment/question. $$z = -3 + i\sqrt{3}$$.
$$r^2 = (-3)^2 + \left(\sqrt{3}\right)^2 = 9 + 3 = 12 \implies$$
$$r = 2\sqrt{3}.$$

Therefore, you are looking for $$\theta$$ such that
$$\cos(\theta) = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$$ and
$$\sin(\theta) = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$$.

Looking at the sign of $$\cos(\theta)$$ and $$\sin(\theta)$$, you immediately deduce that $$\theta$$ is in the 2nd quadrant. Then, for this particular problem, you can simply draw the unit circle, identify the corresponding angle in the first quadrant, and then infer the value of $$\theta$$ in the 2nd quadrant.

For example, you know that $$\cos(\pi/6) = \frac{\sqrt{3}}{2}.$$
Therefore, you can infer, by symmetry, that $$\theta$$ must be $$(\pi - \pi/6) = (5\pi/6)$$.

You could have instead focused on the fact that $$\sin(\pi/6) = (1/2)$$, and reached the same inference. Using symmetry around a drawing of the unit circle is a good way of stretching your intuition to easily handle situations like this.

Anyway, once you determine that $$(\theta = 5\pi/6)$$ and $$(r = 2\sqrt{3})$$, you know that
$$-3 + i\sqrt{3} = 2\sqrt{3}[\cos(5\pi/6) + i\sin(5\pi/6)].$$

For a more systematic way of handling situations like this, first see this wikipedia article.

Then, you could (for example) use the appropriate formula to deduce that $$\sin(\pi - \theta) = \sin(\theta).$$

Then, you can easily attack a situation like this sytematically by first learning (not memorizing) the sine and cosine functions for such special angles as $$\{0, \pi/6, \pi/4, \pi/3, \pi/2\}$$, and the corresponding angles in each of the other 4 quadrants.

Actually, I used the verb learn to indicate that once you've got the 1st quadrant down, you should be able to infer the trig functions that correspond to the special angles in the other quadrants, simply by looking at the unit circle, and making the intuitive leaps, around ideas involving symmetry.

• This wasn't my problem, I don't know how to determine θ. I know If it's in First quadrant, θ is arctan(x/y), but what about other quadrants? Mar 24, 2021 at 21:03
• This is the point that I was making. I was suggesting that you discard the idea that $\theta$ = arctan$(y/x)$ and instead focus on $\theta$ = arccos$(x/r)$ and simultaneously arcsin$(y/r)$. Using this approach, and letting the sign of $(x/r)$ and $(y/r)$ direct you to the proper quadrant, find the angle $\theta$ that simultaneously satisfies both constraints. Again, don't use the tangent function. Use the sine and cosine functions. Mar 24, 2021 at 21:08
• In fact, once you have identified the proper quadrant, you don't actually have to use both the sine and cosine functions. If you know that you are focusing in the proper quadrant, then you can (for example) just satisfy $\cos(\theta) = \frac{x}{r}$. That, in and of itself, plus knowing which quadrant that you are in, will uniquely identify $\theta$. You can of course, adopt the mirror image approach of identifying the quadrant, and then only focusing on $\sin(\theta) = \frac{y}{r}$. Mar 24, 2021 at 21:14
• @someone see my last two comments. Mar 24, 2021 at 21:18
• Okay I know arcsin() and arccos() will give me θ, but in Z=-3+√3i both arcsin() and arccos() are positive numbers. which one should I pick? btw I think I've learned the arctan() procedure. Mar 24, 2021 at 21:48