I need to take a raincheck with this problem. I want to make sure I haven't messed up some fundamental idea.
Convert the complex number $$-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i$$ to polar form.
I took the modulus as below,
$$\lvert-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i \rvert = \sqrt{(\dfrac{-1}{2})^2 + (\dfrac{\sqrt 3}{2})^2} = 1$$
And the argument as below,
$$arg(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1}) = -60 = -\dfrac{\pi}{3}$$
Hence the complex number in polar form is,
$$-\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}$$
But, The required answer is $$\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}$$
I thought of converting the -60 to positive, as 360 – 60 = 300, ie:- $$\dfrac{5\pi}{3}$$. I have a feeling I am missing something important. Can you guys tell me where I am going wrong? Thanks for all your help!
Best Answer
The problem is that points can be expressed in polar form in more than one way. Take a look at the diagram below:
There's your point (in blue). As you can see, it's in the second quadrant, and the angle $\theta = \frac{2\pi}{3}$ passes through it. So one could accurately say that the point's polar representation is $r = 1, \theta = \frac{2\pi}{3}$, i.e. $$z = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)$$
You'll notice, though, that $\theta = -\frac{\pi}{3}$ represents the same angle, but in the opposite direction. With this in mind, one could just as accurately say that the point's polar representation is $r = -1, \theta = -\frac{\pi}{3}$, i.e. $$z = -\cos\left(-\frac{\pi}{3}\right) - i\sin\left(-\frac{\pi}{3}\right)$$
Because $\cos\left(\frac{2\pi}{3}\right) = -\cos\left(-\frac{\pi}{3}\right)$ and $\sin\left(\frac{2\pi}{3}\right) = -\sin\left(-\frac{\pi}{3}\right)$, these two polar representations are equivalent. In other words, there's nothing wrong with concluding, as you did, that $\theta = -\frac{\pi}{3}$ You and the so-called "required answer" can reasonably disagree here and both be correct.
But you still did something wrong. If you choose to use $\theta = -\frac{\pi}{3}$, then you must also choose $r = -1$, yet you choose $r = 1$, based on your modulus computation. How can you avoid making this mistake in the future?
Simple: look at the quadrant in which the point lies. The point $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ has negative real part and positive imaginary part. Therefore it must be in the second quadrant. The angle $\theta = -\frac{\pi}{3}$ extends into the fourth quadrant, so if you intend to use that angle, then you must also make $r$ negative.