How will I prove arg (z1/z2)=arg z1-arg z2
K kidia New member Joined Apr 11, 2006 Messages 27 Apr 11, 2007 #1 How will I prove arg (z1/z2)=arg z1-arg z2
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Apr 11, 2007 #2 \(\displaystyle \L \begin{array}{rcl} \frac{z}{w} & = & \frac{{z\bar w}}{{\left| w \right|^2 }} \\ z & = & \left| z \right|cis(\arg (z))\quad \& \quad \bar w = \left| w \right|cis(\arg (\bar w)) \\ \arg (\bar w) & = & - \arg (w) \\ z\bar w & = & \left[ {\left| z \right|cis(\arg (z))} \right]\left[ {\left| w \right|cis( - \arg (w))} \right] \\ & = & \left[ {\left| z \right|\left| w \right|cis(\arg (z) - \arg (w))} \right] \\ \mbox{so} \\ \arg \left( {\frac{z}{w}} \right) & = & \arg (z) - \arg (w) \\ \end{array}\)
\(\displaystyle \L \begin{array}{rcl} \frac{z}{w} & = & \frac{{z\bar w}}{{\left| w \right|^2 }} \\ z & = & \left| z \right|cis(\arg (z))\quad \& \quad \bar w = \left| w \right|cis(\arg (\bar w)) \\ \arg (\bar w) & = & - \arg (w) \\ z\bar w & = & \left[ {\left| z \right|cis(\arg (z))} \right]\left[ {\left| w \right|cis( - \arg (w))} \right] \\ & = & \left[ {\left| z \right|\left| w \right|cis(\arg (z) - \arg (w))} \right] \\ \mbox{so} \\ \arg \left( {\frac{z}{w}} \right) & = & \arg (z) - \arg (w) \\ \end{array}\)