This is a problem in my undergrad foundations class.
\begin{equation}
w^2=-\sqrt{3} - i \\w=(-\sqrt{3}-i)^{\frac{1}{2}} \\w=\sqrt{2}\bigg(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\bigg)
\end{equation}
So I get to here and the next step is
\begin{equation}
\sqrt{2}\bigg(\cos\bigg(\frac{5\pi}{12}+\pi\bigg)+i\sin\bigg(\frac{5\pi}{12}+\pi\bigg)\bigg)
\end{equation}
and this is equal to
\begin{equation}
w(\cos\pi+i\sin\pi)=-w
\end{equation}
Can someone help me with understanding the last two steps. Thanks
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Asaf Karagila
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Lanous
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you dropped a sign at near the beginning. – Doug M Mar 03 '17 at 18:16
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1Hint: $\cos (\theta + \pi) = -\cos \theta$ and $\sin (\theta + \pi) = -\sin \theta$ for all $\theta$. – Connor Harris Mar 03 '17 at 18:21
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The step before last has an extra $-ve$ sign than the step before it. – Emptymind Mar 03 '17 at 18:22
1 Answers
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$\omega^2 = $$2(-\frac {\sqrt 3}{2} + i \frac 12)\\ 2(\cos \frac {7\pi}{6} + i \sin \frac {7\pi}{6})$
$\omega = \sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12})$
And here you are....
When we take the square root of the square of something, there are always two solutions. For example $x^2 = 9 \implies x = \pm 3$
We can put in the $\pm symbol here, but that technique breaks down with higher roots.
A more general approach...
$\omega^2 =2(\cos (\frac {7\pi}{6}+2n\pi) + i \sin (\frac {7\pi}{6}+2n\pi))$
Takes advantage of the period nature of the trig functions.
$\omega =\sqrt 2(\cos (\frac {7\pi}{12}+n\pi) + i \sin (\frac {7\pi}{12}+n\pi))$
And now we can look at just the solutions where the argument is in $[0,2\pi)$
$\omega =\sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12}),\sqrt 2(\cos (\frac {7\pi}{12}+\pi) + i \sin (\frac {7\pi}{12}+\pi))$
Doug M
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So because of the period nature the second solution will always be $+n\pi$ and therefore $\sqrt 2(\cos (\frac {7\pi}{12}+\pi) + i \sin (\frac {7\pi}{12}+\pi))$ is $-w$ in your example? – Lanous Mar 03 '17 at 18:44
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