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GRE Subject Test: Math : Complex Plane

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Example Questions

Example Question #111 : Classifying Algebraic Functions

Write the complex number $1-\sqrt{3}i$ in polar form, where polar form expresses the result in terms of a distance from the origin on the complex plane and an angle from the positive -axis, $\theta$ , measured in radians.

Possible Answers:

$\newline r=\sqrt{3} \newline \newline \theta =2 \pi$

$\newline r=\frac{1}{\sqrt{3}} \newline \newline \theta = \frac{3 \pi}{5}$

$\newline r=2 \newline \newline \theta = \frac{5 \pi}{6}$

$\newline r=1 \newline \newline \theta = \frac{\sqrt{3} \pi}{6}$

$\newline r=2 \newline \newline \theta = \sqrt{3} \pi$

Correct answer:

$\newline r=2 \newline \newline \theta = \frac{5 \pi}{6}$

Explanation:

To see what the polar form of the number is, it helps to draw it on a graph, where the horizontal axis is the imaginary part and the vertical axis the real part. This is called the complex plane.

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To find the angle $\theta$ , we can find its supplementary angle $\Psi$ and subtract it from $\pi$ radians, so $\theta = \pi - \Psi$ .

Using trigonometric ratios, $\tan \Psi = \frac{1}{\sqrt{3}}$ and $\arctan({\frac{1}{\sqrt{3}}})= \Psi = \frac{\pi}{6}$ .

Then $\theta = \frac{5\pi}{6}$ .

To find the distance , we need to find the distance from the origin to the point $(-\sqrt{3}, 1)$ . Using the Pythagorean Theorem to find the hypotenuse , $r = \sqrt{ \sqrt{3}^2 + 1^2 }$ or r=2 .

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GRE Subject Test: Math : Complex Plane

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #111 : Classifying Algebraic Functions

All GRE Subject Test: Math Resources

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