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How can I calculate the argument of the complex number $z= (\frac{1}{2}+ \frac{i\sqrt{3}}{2}) \cdot (1+i)$?

I always get $\tan^{-1}(-2-\sqrt{3})$, but the book answer is $7 \pi/12$.

NoChance
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3 Answers3

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You're given the two numbers as a multiplication. Recall that $r_1e^{i\theta} \cdot r_2e^{i \phi} = r_1r_2e^{i(\theta + \phi)},$ which is to say that arguments get added together when multiplying.

Now, note that $\frac12 + \frac{\sqrt3}2$ has argument $\frac\pi3$, while $1 + i$ has argument $\frac\pi4$. The argument of their product, then, is $\frac\pi3 + \frac\pi4 = \frac{7\pi}{12}$.

You're not wrong when you say that the argument is $\tan^{-1}(-2 - \sqrt{3})$, because

\begin{align*} \tan \frac{7\pi}{12} &= \tan \left(\frac\pi3 + \frac\pi4\right) \\ &= \frac{\sqrt{3} + 1}{1 - (\sqrt{3})(1)} \\ &= \ldots \\ &= -2 - \sqrt{3} \end{align*} using the identity for $\tan(\alpha + \beta)$. It's just that it's not at all obvious that $\tan^{-1}(-2 - \sqrt{3}) = \frac{7\pi}{12}$, unlike for nicer angles whose tangents you've probably memorized!

So in this case, the "arguments get added" approach is the way to go.

pjs36
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  • It's not obvious that $\arctan(-2-\sqrt 3)$ is ... anything. But the square root makes me think its a "nice" ratio probably a rational multiple of pi. So calculator time: arctan(-2-3root(3) = -1.3089969389957471826927680763665. And -1.3089969389957471826927680763665/pi = -0.41666666666666666666666666666667 which is clearly something divided by 3. Easy to see it is -5/12 so the argument is $-5\pi/12$. Which in positive terms is $7\pi/12$. – fleablood Mar 24 '16 at 17:08
  • Yeah, that's a fair point -- the $\sqrt{3}$ does indicate it might have something to do with $\pi/3$ in some capacity. – pjs36 Mar 24 '16 at 17:17
  • It's not so much that $\sqrt 3$ indicate $\pi/3$ but that $\sqrt{n}$ indicates some sort of right triangle with integer sides... maybe. ($radius= r, rcos = n \implies rsin = \sqrt{r^2 - n^2}$). In any event when you get a solution artan (whatever) it doesn't hurt to calculate it out, divide by pi and see if you get so nice rational*pi solution. – fleablood Mar 24 '16 at 17:28
  • On a side note. I kind of wish calculators had a single "display as $a\pi$" button or display mode. It's kind of pointless to calculate in radians and to give a decimal value rather than an $xpi$ value. $.4166666666667\pi$ would be a LOT* more useful than 1.3089969389957471826927680763665 – fleablood Mar 24 '16 at 17:34
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It is $\arctan(\sqrt{3})+\arctan(1) = \frac{\pi}{3}+\frac{\pi}{4}$.

  • How do you mean? My book says that the angle is 7 pi/12 –  Mar 24 '16 at 16:17
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    The answer is $\frac{\pi}{4}$ for the problem as it was originally posted when Hamid answered, as the $i$ was missing from the first factor. –  Mar 24 '16 at 16:22
  • That's true. I didn't mention the i but now it is edited –  Mar 24 '16 at 16:28
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I) Tedious: $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i)= (1/2 - \sqrt{3}/2) + (1/2 + \sqrt 3/2) i = $. Arg($z$) = $\arctan (1+ \sqrt{3})/(1-\sqrt{3}) = -5\pi/12 = 7\pi/12$.

(To be fair, I made data entry errors the first six times I tried to enter this into a calculator.)

II) Easier (albeit it notation intimidating): $\arg (1+i) = \pi/4$ so $i+1 = r*e^{\frac{\pi}4 i}$. ($r = \sqrt 2$ but we don't give a toss.) $\arg (1/2 + i\sqrt{3}/2) = \pi/3$ so $1/2 + i\sqrt{3}/2 = s*e^{\frac{\pi}3 i}$. ($s = 1$ but we don't give a toss.)

So $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i) = r*e^{\frac{\pi}4 i}*s*e^{\frac{\pi}3 i} = rse^{(\pi/3 + \pi/4)i}$ so $\arg(z) = \pi/3 + \pi/4 = 7\pi/12$.

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Post Script: $\frac {1 + \sqrt 3}{1 - \sqrt 3}= \frac{(1 + \sqrt 3)^2}{1 - 3} = \frac {4 + 2\sqrt 3}{-2} = -2 - \sqrt 3$ so your first answer was absolutely correct. Why did you doubt yourself? Why did you assume $\tan^{-1}(-2 - \sqrt{3}) \ne 7\pi/12$?

fleablood
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  • I used a calculator. Then I divided by pi to see if it had a "nice" answer. (Actually, I figured the answer by adding the args. Kept trying to use the calculator, made data entry errors six times and kept trying till I got the answer I wanted). BTW. If sin/cos = -2-\sqrt3 then there is a right triangle with a=1, b=2+\sqrt3, h = 8 + 4\sqrt3, so this is a triangle with 1 side = 1/4 the hypotenuse. There's probably some geometry to get that this means angles of pi/12 and 5pi/12s but I can't figure it out off the top of my head. – fleablood Mar 24 '16 at 18:11