## Precalculus (6th Edition) Blitzer

The rectangular form of the given complex number is $4\sqrt{2}-4\sqrt{2}i$.
Consider any complex number, given by $z=x+iy$, for a complex number in rectangular form, $z=x+iy$ …… (1) The polar form is given by, $z=r\left( \cos \theta +i\sin \theta \right)$ …… (2) Here, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta$. Divide the value of y by x, to get, $\tan \theta =\frac{y}{x}$ Also, the value of r is called the moduli of the complex number, given by, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ For any complex number in polar form, $z=r\left( \cos \theta +i\sin \theta \right)$, the rectangular form is, Using (1) and (2), \begin{align} & z=8\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right) \\ & z=x+iy \\ \end{align} Simplify it further to get, \begin{align} & 8\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)=8\cos \frac{7\pi }{4}+8i\sin \frac{7\pi }{4} \\ & =\left( 8\times \frac{1}{\sqrt{2}} \right)+i\left( 8\times -\frac{1}{\sqrt{2}} \right) \\ & =4\sqrt{2}-4\sqrt{2}i \\ & 8\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)=4\sqrt{2}-4\sqrt{2}i \end{align} The rectangular form of the complex number is $4\sqrt{2}-4\sqrt{2}i$. The rectangular form of the given complex number is $4\sqrt{2}-4\sqrt{2}i$.