+0

# to polar form

0
62
2
+35

https://prnt.sc/s6h4k1

Apr 27, 2020

#1
+24995
+2

to polar form

$$\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{12\sqrt{2}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {3\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} } \\\\ &=& \dfrac{4\sqrt{2}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{2}\sqrt{3}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {\sqrt{3}\sqrt{3}\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6}\mathop{cis}\left(\dfrac{7\pi}{6}\right)} {3\mathop{cis}\left(\dfrac{\pi}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6} e^{i\frac{7\pi}{6}} } {3 e^{i\frac{\pi}{4}} } \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{7\pi}{6}}e^{-i\frac{\pi}{4}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{7\pi}{6}-i\frac{\pi}{4}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi\left(\frac{7}{6}- \frac{1}{4}\right)} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \left(\frac{28-6}{24} \right)} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \frac{22}{24}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\pi \frac{11}{12}} \\\\ &=& \dfrac{4\sqrt{6}}{3} e^{ i\frac{11\pi}{12}} \\\\ &=& \mathbf{\dfrac{4\sqrt{6}}{3} \mathop{cis}\left( \dfrac{11\pi}{12}\right)} \\ \hline \end{array}$$

Apr 27, 2020
edited by heureka  Apr 27, 2020
#2
+21017
+1

There is another way to do this:  since it is a division problem in cis form divide the constants and subtract the angles.

[ 12 · sqrt(2) ] / [ 3 · sqrt(3) ]  =  [ 4 · sqrt(2) ] / sqrt(3)  =  [ 4 · sqrt(6) ] / 3

[ cis( 7·pi/6 ) ] / [ cis( pi/4 ) ]  =   cis( 7·pi/6 - pi/4 )  =  cis( 14·pi/12 - 3·pi/12 ]  =  cis( 11·pi/12 )

Answer:  [ 4 · sqrt(6) ] / 3 · cis( 11·pi/12 )

Apr 27, 2020