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

MATH152 April 2016

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Question B 05 (a)
Consider $z=-1/2+i{\sqrt {3}}/2$ . Recall that $\tan ^{-1}{\sqrt {3}}=\pi /3$ . (a) Mark the approximate location of $z$ in a sketch of the complex plane.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
For $z=a+ib$ : $a,b$ correspond to $x,y$ values in the plane.

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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. $z=-1/2+i{\sqrt {3}}/2$ : $a=-1/2,b={\sqrt {3}}/2$ , it has negative x-value and positive y-value so it should be in the second quadrant. Suppose the angle between $z$ and $x$ -axis is $\theta$ , we have $\tan \theta ={\frac {b}{a}}=-{\sqrt {3}}$ . It's known $\tan x=-\tan(\pi -x)$ and $\tan(\pi /3)={\sqrt {3}}$ , $\theta \in [90^{\circ },180^{\circ }]$ . Thus the angle between $z$ and $x$ -axis is ${\frac {2\pi }{3}}$