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!

  • $\begingroup$ The problem is with arctanget. The Wikipedia article atan2 is worth reading. $\endgroup$ – Henry May 26 '11 at 10:21

The problem is that points can be expressed in polar form in more than one way. Take a look at the diagram below: polar graph of point

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.

  • $\begingroup$ Thank you! That was a great explanation. I have one question regarding, r = -1. So far, I have been discarding the negative root of r assuming that r must be positive as it is a modulus operation. Can you clarify this case? $\endgroup$ – mathguy80 May 26 '11 at 11:49
  • $\begingroup$ @mathguy80: There's a question here what is meant by "a/the polar representation" of a complex number. @yunone and I assumed that by this you meant the modulus/argument representation. The modulus is (as you say) defined to be positive, and the argument is defined as the angle formed with the $x$-axis, not its opposite. @Alex, on the other hand, is using "polar representation" to mean any representation of the form $z=r\mathrm e^{\mathrm i\phi}$ with real values of $r$ and $\phi$. In that case, you can use either the modulus and the argument, or minus the modulus and the argument plus $\pi$. $\endgroup$ – joriki May 26 '11 at 12:27
  • $\begingroup$ Ah, I see what you mean about different representation. Learned some cool stuff and was able to plug some holes in my understanding today. Thanks! $\endgroup$ – mathguy80 May 26 '11 at 16:15
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    $\begingroup$ What did you use to make that image? $\endgroup$ – davidlowryduda May 26 '11 at 21:58
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    $\begingroup$ @mixedmath: Grapher. It's a free 2D and 3D graphing app that comes bundled with every Mac. Most Mac users don't even know they have it, though, because Apple buries it in the Utilities folder, which is itself in the Applications folder. If you have a Mac, you should definitely check it out. Go to the Examples menu to see some of what it can do. $\endgroup$ – Alex Basson May 26 '11 at 22:02

If you draw the picture, it often points you in the right direction. You can see in the picture how the 30-60-90 triangle is oriented and you will get it right every time.

  • $\begingroup$ That's a nice way to remember this. Thanks. $\endgroup$ – mathguy80 May 29 '11 at 15:18

The problem is in how you calculated the argument using the arctangent. The range of the arctangent is only $(-\pi/2,\pi/2)$, so you can't get the full range of arguments in this way. Your number is in the second quadrant, which yields the same tangent values as the fourth quadrant, and the argument you calculated is in the fourth quadrant. You need to add $\pi$ to get the right argument.

There's another, unrelated error: The cosine is even and the sine is odd, so the negative sign of the argument you calculated should have appeared in front of the sine but not the cosine; i.e., the polar form corresponding to the argument you calculated is

$$\cos\frac{\pi}{3}-\mathrm i\sin\frac{\pi}{3}\;.$$

  • $\begingroup$ Good stuff about arctangent. And you are right about the mistake in the sign. $\endgroup$ – mathguy80 May 26 '11 at 11:51

The problem is you forgot to add $\pi$ in calculating the argument. See the wiki page which shows all the formulas for computing the argument depending on $x$ and $y$.

Taking $z=x+yi$ to be your complex number, here $x$ is negative and $y$ is positive, so $$ \text{arg}(-\dfrac{1}{2} + \dfrac{\sqrt3}{2}i) = \tan^{-1} (\dfrac{\sqrt 3}{2} \times \dfrac{-2}{1})+\pi = -\dfrac{\pi}{3}+\pi=\frac{2\pi}{3}. $$


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