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This is the final question from IB Math HL Paper 1 Nov 2014 TZ0.

Show that: $$(1+ i\tan\theta)^n + (1-i\tan\theta)^n = \frac{2\cos n\theta}{\cos^n\theta}$$

I tried to solve the problem as follows:

We know: $$r=\sqrt{x^2+y^2}$$

I started off by converting the left side to polar form in order to use De Moivre's Theorem.
So we get: $$r=\sqrt{\tan^2\theta+1}$$ $$r=\sqrt{\sec^2\theta}$$ $$r=|\sec\theta|$$

Thus:
$$(\frac{1}{|(cos\theta)|}(\cos\theta+i\sin\theta))^n + (\frac{1}{|(\cos\theta)|}(\cos\theta-i\sin\theta))^n$$ By De Moivre's Theorem:

$$(\frac{1}{|(\cos\theta)|^n}(\cos n\theta+i\sin n\theta)) + (\frac{1}{|(\cos\theta)|^n}(\cos n\theta-i\sin n\theta))$$

Simplifying we get:

$$\frac{2\cos n\theta}{|\cos\theta|^n}$$

Which is the right hand side if we ignore the absolute value. But why are we supposed to ignore the absolute value as $\theta$ has no restrictions except for $\cos\theta\neq0$?

I would think my answer with the absolute value is correct, but testing with $\theta = \pi$ and $n=3$, we get $2 = -2$, which is not true. Testing without the absolute bracket we get $2 = 2$, which holds true.

So we know my answer does not work, but my question is why should we simply ignore the absolute value on $\sec\theta$?

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  • $\begingroup$ Your value of "$r$" is correct, but your angle "$\theta$" is wrong; shouldn't it be $\tan^{-1}(\tan\theta)$? This will deal with the signs, I think. But as the answer below shows, it is not how to do it, just use algebra to take out the $\cos\theta$. $\endgroup$ – ancientmathematician Jul 3 '18 at 10:01
  • $\begingroup$ won't that just leave $\theta$ as the angle? $\endgroup$ – Ryan Liu Jul 3 '18 at 17:50
  • $\begingroup$ Not always, we've got to look at the signs of $\cos\theta$ and $\sin\theta$ to choose the correct angle, see @somos 's solution. $\endgroup$ – ancientmathematician Jul 4 '18 at 11:13
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$$(1\pm i\tan t)^n=\dfrac{(\cos t\pm i\sin t)^n}{\cos^nt}=\dfrac{\cos nt\pm i\sin nt}{\cos^nt}$$

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  • $\begingroup$ But why is there no absolute value on the $\cos t$ in the denominator? $\endgroup$ – Ryan Liu Jul 3 '18 at 9:55
  • $\begingroup$ @RyanLiu, Assumed $n$ to be an integer $\endgroup$ – lab bhattacharjee Jul 3 '18 at 10:15
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You are confused about two angles. First, use the polar form for complex numbers. That is, let $$\ r e^{\ i \ \phi} = r ( \cos \phi + i \sin \phi ) = r \cos \phi + i \ r \sin \phi = 1 + i \tan \theta. \tag{1} \ $$ Now, $\ r^2 = 1 + \tan^2 \theta. \ $ Solving for $\ r \ $ gives $\ r = |\sec \theta|, \ $ and then substituting it in equation $(1)$ while equating real and imaginary parts gives $\, |\sec \theta|\, \cos \phi = 1, \,$ and $\ |\sec \theta|\ \sin \phi = \tan \theta. \ $ This implies $\ |\cos \theta| = \cos \phi \ $ and $\ \tan \theta = \tan \phi. \ $ Thus, either $\ \cos \theta = \cos \phi \ $ and $\ \theta = \phi, \ $ or else $\ \cos \theta = - \cos \phi \ $ and $\ \theta = \pi - \phi. \ $

Finally, since $\ r\ e^{\ i \ \phi} = 1 + i \tan \theta ,\ $ then raising to power $n$ gives $\ r^n e^{\ i\, n\ \phi} = (1 + i \tan \theta)^n, \ $ but now $\ r^n \cos (n\phi) = \cos (n \phi)/|\cos \theta|^n = \cos (n \theta)/(\cos \theta)^n. \ $ In the case $\ |\cos \theta| = \cos \theta = \cos \phi, \ $ we have $\ \phi = \theta \ $ and the equation is an identity. In the other case $\ |\cos \theta| = - \cos \theta = \cos \phi, \ $ and $\ \cos(n \phi) / (\cos \phi)^n = \cos (n \theta)/(\cos \theta)^n. \ $ Depending on the parity of $\ n \ $ the numerators and denominators of both sides of the equation are the same or differ in sign. The gist is that we do not "ignore the absolute value on $\sec\theta$" but use the polar form of $\ 1 + i \tan\theta \ $ using another angle.

After all this, another answer shows that it is not needed. That is because you get directly that $\ \cos \theta \ ( 1 \pm i\tan\theta) = \cos \theta \pm i\sin\theta \ $ and this gives you a simple answer without polar form.

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  • $\begingroup$ Thanks for the response, but I am confused on where does $|\sec\theta|\cos\theta = 1$, $|\sec\theta|\sin\theta = \tan\theta$, and $\phi=\pi-\theta$ come from? $\endgroup$ – Ryan Liu Jul 3 '18 at 18:09

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