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Expert-verified Found in: Page 591 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # In Problems $11-22$, plot each complex number in the complex plane and write it in polar form. Express the argument in degrees.$-2+3i$.

The polar form of $-2+3i$ is, $\sqrt{13}\left(\mathrm{cos}123.{7}^{\circ }+i\mathrm{sin}123.{7}^{\circ }\right)$.

The given complex number is plotted in the complex plane. See the step by step solution

## Step 1 Given argument is,

$-2+3i$.

The rectangular co-ordinates of the point $\left(-2+3i\right)$ is, $\left(-2,3\right)$.

## Step 2 The point is plotted in the complex plane. From the graph it is clear that the point is located in the first quadrant.

## Step 3 We know that,

$\left|z\right|=r\phantom{\rule{0ex}{0ex}}r=\sqrt{{x}^{2}+{y}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{{\left(-2\right)}^{2}+{3}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{13}$

Use $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$ to find the value of $\theta$.

$\mathrm{sin}\theta =\frac{y}{r}\phantom{\rule{0ex}{0ex}}=\frac{3}{\sqrt{13}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{x}{r}\phantom{\rule{0ex}{0ex}}=\frac{-2}{\sqrt{13}}$

Since $\mathrm{sin}\theta$ is $\frac{3}{\sqrt{13}}$ and $\mathrm{cos}\theta =\frac{-2}{\sqrt{13}}$,the value of $\theta$ is, $123.{7}^{\circ }$.

## Step 4 The polar form of z=x+iy is, rcos θ+i sin θ.

The polar form of $-2+3i$ is,$\sqrt{13}\left(\mathrm{cos}123.{7}^{\circ }+i\mathrm{sin}123.{7}^{\circ }\right)$. ### Want to see more solutions like these? 