Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 05 (c)

MATH152 April 2016

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Question B 05 (c)
Consider $z=-1/2+i{\sqrt {3}}/2$. Recall that $\tan ^{-1}{\sqrt {3}}=\pi /3$. (c) Write $z$ in polar form. That is, find a real number $r>0$ and $0\leq \theta <2\pi$ such that $z=re^{i\theta }$.

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Hint
In the polar form of a complex number $z=a+bi=re^{i\theta }$, two variables $r$ and $\theta$ are determined by the relations $a=r\cos \theta$ and $b=r\sin \theta$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For any complex number $z=a+ib=r(\cos \theta +\sin \theta )=re^{i\theta }$, we have $r=|z|={\sqrt {a^{2}+b^{2}}}$ and $\tan \theta ={\frac {b}{a}}$. We know from part (b) that $r=|z|=1.$ We also have $\tan \theta ={\frac {\frac {\sqrt {3}}{2}}{-{\frac {1}{2}}}}=-{\sqrt {3}}$. We are given that $\tan ^{-1}{\sqrt {3}}=\pi /3$. From part (a), $z$ is in the second quadrant, which is consistent with a negative tangent. So we know that $\theta$ must form an angle of $\pi /3$ with the $x$-axis and lie in the second quadrant. We also require that $0\leq \theta <2\pi$. Hence $\theta ={\frac {2}{3}}\pi .$ Therefore, the desired polar form is $\color {blue}z=e^{{\frac {2}{3}}\pi i}$ .

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