Question

In Problems $11-22$, plot each complex number in the complex plane and write it in polar form. Express the argument in degrees.

$-2+3i$.

Short answer

The polar form of $-2+3i$is, $\sqrt{13}\left(\cos 123.7^{\circ }+i\sin 123.7^{\circ }\right)$.

The given complex number is plotted in the complex plane.

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,

$$\begin{gathered} \left|z\right|=r \\ r=\sqrt{x^2+y^2} \\ r=\sqrt{{\left(-2\right)}^2+3^2} \\ r=\sqrt{13} \end{gathered}$$

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

$$\begin{gathered} \sin \theta =\frac{y}{r} \\ =\frac{3}{\sqrt{13}} \\ \cos \theta =\frac{x}{r} \\ =\frac{-2}{\sqrt{13}} \end{gathered}$$

Since $\sin \theta$is $\frac{3}{\sqrt{13}}$and $\cos \theta =\frac{-2}{\sqrt{13}}$,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,$\sqrt{13}\left(\cos 123.7^{\circ }+i\sin 123.7^{\circ }\right)$.