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Given that $z_1=2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)$ and $z_2=2\operatorname{cis}\left(\frac{2\pi}{3}\right)$ find the smallest positive value of $n$ such that $\left(\frac{z_1}{z_2}\right)^n \in \Bbb{R}^+$.

My attempt:

$$\frac{z_1}{z_2}=\frac{2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)}{2\operatorname{cis} \left(\frac{2\pi}{3}\right)}=\sqrt{3}\operatorname{cis}\left(\frac{5\pi}{6}\right)$$

Hence,

$$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^n\operatorname{cis}\frac{5n\pi}{6}$$

Since $(\frac{z_1}{z_2})^n \in \Bbb{R}^+$, so

$$\sqrt{3}^ni\sin\frac{5n\pi}{6}=0$$ $$5n\pi=0,6\pi$$ $$n=0,1.2$$

At the same time, I also know that

$$\sqrt{3}^n\cos\frac{5n\pi}{6}>0$$ Solving for the above will lead to $n>\frac{3}{5}$

However, the textbook answer is $n=12$.

Somehow, I could not reconcile the my two solutions to $n=12$.

May I know where did I did wrongly? Thank you in advance.

IM_LOST
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    I conjecture that the question was meant to be smallest POSITIVE INTEGER value of n. – insipidintegrator Feb 24 '23 at 06:51
  • @insipidintegrator Even if it is the smallest positive integer, how do I get $n=12$? I thought $n=6$ is the smallest positive integer. – IM_LOST Feb 24 '23 at 06:55
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    Use \sin and \cos to get the proper font and spacing for $\sin$ and $\cos$. For functions like $\operatorname{cis}$ that don't have a command of their own, you can use \operatorname{cis}. Also, you can get properly sized parentheses (and other paired delimiters) that adapt to their content by preceding them with \left and \right. – joriki Feb 24 '23 at 07:02
  • @joriki Done and thank you for pointing out. – IM_LOST Feb 24 '23 at 07:16
  • If you right-click on some formatted text, then hover over "Show Math As" and slide right to "Tex Commands", and then left-click, you can see the LaTex/MathJax code. I find this useful in learning the coding, (and copying), and often much easier than trying to find a code in a huge list of symbols . E.g. $\nabla$ is called "del" in multi-variable calculus but its code is \nabla . I dk why. – DanielWainfleet Feb 24 '23 at 07:57

2 Answers2

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For $n\in\Bbb Z$ we have

$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^{\,n}\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$

iff $\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$

iff $(\,\cos \frac{5n\pi}{6}>0 \, \land \,\sin \frac{5n\pi}{6}=0\,)$

iff $(\,\cos \frac{5n\pi}{6}=1 \, \land \,\sin \frac{5n\pi}{6}=0\,)$

iff $\frac {5n}{6}\,$ is twice an integer

iff $\,12|n$.

DanielWainfleet
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Your mistake is in “Solving the above will lead to $n\gt\frac35$”. The cosine oscillates; you can’t convert an inequality for the cosine into a simple inequality for its argument like that. (Actually, even if you did it locally, it should have been $n\lt\frac35$.) If you substitute $n=6$, you get $\cos5\pi=-1$.

joriki
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  • I am still confused. Would you be able to share a little more on the steps to the solution? – IM_LOST Feb 24 '23 at 07:19
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    He means to say that $\cos(x) > 0$ does not imply $x>\frac{\pi}{2}$. Instead consider all the intervals where $\cos(x)>0$ and then choose minimum $n$ from there. – Sandipan Samanta Feb 24 '23 at 07:45