N Naeema F New member Joined Jun 11, 2020 Messages 8 Oct 9, 2020 #1 I have calculated the polar form of w = i - √3 to be 2cis(7/6 π) Please confirm if this is correct. Much appreciated.
I have calculated the polar form of w = i - √3 to be 2cis(7/6 π) Please confirm if this is correct. Much appreciated.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,760 Oct 9, 2020 #2 \(\displaystyle |w|= \sqrt{1^2+ (-\sqrt{3})= \sqrt{1+ 3}= \sqrt{4}= 2\) and \(\displaystyle arg(w)= arctan(-\frac{1}{\sqrt{3}})=arctan(-\frac{\sqrt{3}}{3})=-\frac{6}{7}\pi\). Yes, that is correct.
\(\displaystyle |w|= \sqrt{1^2+ (-\sqrt{3})= \sqrt{1+ 3}= \sqrt{4}= 2\) and \(\displaystyle arg(w)= arctan(-\frac{1}{\sqrt{3}})=arctan(-\frac{\sqrt{3}}{3})=-\frac{6}{7}\pi\). Yes, that is correct.
