## Precalculus (6th Edition) Blitzer

The division of the complex numbers in the polar form is $2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)$.
Consider the given complex number to write in the polar form, \begin{align} & {{z}_{1}}=2+2i \\ & {{z}_{2}}=1+i \\ \end{align} (I) For a complex number $z=x+iy$, the polar form is given by, $z=r\left( \cos \theta +i\sin \theta \right)$ (II) Here, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta$, dividing the value of y by x, to get $\tan \theta =\frac{y}{x}$ (III) Also, the value of r is called the moduli of the complex number, given by, For ${{z}_{1}}=2+2i$ \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & r=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\ & r=\sqrt{4+4} \\ & r=2\sqrt{2} \\ \end{align} $r=2\sqrt{2}$ (IV) For any complex number $z=x+iy$, the sign of the value of x and y determine in which quadrant the value of $z=x+iy$ would lie, If the value of x is positive and the value of y is positive, then the angle $\theta$ lies in the first quadrant having the value of $\theta$ as $\theta$ (V) Also, if the value of x is negative and the value of y is positive, then the angle $\theta$ lies in the second quadrant having the value of $\theta$ as $\pi -\theta$ (VI) Also, if the value of x is negative and the value of y is negative, then the angle $\theta$ lies in the third quadrant having the value of $\theta$ as $\pi +\theta$ (VII) Also, if the value of x is positive and the value of y is negative, then the angle $\theta$ lies in the fourth quadrant having the value of $\theta$ as $2\pi -\theta$ (VIII) For the given complex number, using (I), (III) to get, \begin{align} & \tan \theta =\frac{y}{x} \\ & \tan \theta =\frac{2}{2} \\ & \tan \theta =1 \\ \end{align} For the given complex number, Using (V), $\theta =\frac{\pi }{4}$ (IX) Using (II), (IV), and (IX), to get, ${{z}_{1}}=2\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)$ The polar form of the complex number is ${{z}_{1}}=2\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)$ For ${{z}_{2}}=1+i$ \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & r=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\ & r=\sqrt{1+1} \\ & r=\sqrt{2} \\ \end{align} $r=\sqrt{2}$ (X) For the given complex number, using (I), (III) to get, \begin{align} & \tan \theta =\frac{y}{x} \\ & \tan \theta =\frac{1}{1} \\ & \tan \theta =1 \\ \end{align} For the given complex number, Using (VIII), $\theta =\frac{\pi }{4}$ (XI) Using (II), (IV), and (IX), to get, ${{z}_{2}}=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)$ The polar form of the complex number is ${{z}_{2}}=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)$ Consider any complex number, given by, \begin{align} & {{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\ & {{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\ \end{align} For a complex number in polar form, the division is calculated as, $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)$ (XII) The polar form after the division of the complex numbers, \begin{align} & {{z}_{1}}=2\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\ & {{z}_{2}}=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\ \end{align} (XIII) Divide it using (XII) and (XIII), to get, \begin{align} & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \\ & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{2\sqrt{2}}{\sqrt{2}}\left( \cos \left( \frac{\pi }{4}-\frac{\pi }{4} \right)+i\sin \left( \frac{\pi }{4}-\frac{\pi }{4} \right) \right) \\ & =2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right) \end{align} The division of the complex numbers in the polar form is $2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)$