## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Review Exercises - Page 410: 42

#### Answer

For the absolute value of $z = 2$, the graph of all the complex numbers $z$ will have their vector vertices lying on the circle $x^2 + y^2 = 2^2$ or on the circle with center at the origin and radius = 2.

#### Work Step by Step

For $z = x + yi$, the absolute value of $z$ is $\sqrt{x^2 + y^2}$. Now, for the absolute value of $z = 2$, $\sqrt{x^2 + y^2} = 2$ or $x^2 + y^2 = 2^2$ which is a circle with center at the origin and radius = 2. The graph of all the complex numbers $z$ will have their vector vertices lying on the circle $x^2 + y^2 = 2^2$ or on the circle with center at the origin and radius = 2.

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