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kidia
04-11-2007, 05:27 PM
How will I prove arg (z1/z2)=arg z1-arg z2

pka
04-11-2007, 05:52 PM
\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}