What would the magnitude of e^(2-3i) be? Since it is in the 3rd quadrant, what is its argument?
L lastlydreaming New member Joined Oct 7, 2006 Messages 10 Oct 7, 2006 #1 What would the magnitude of e^(2-3i) be? Since it is in the 3rd quadrant, what is its argument?
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Oct 7, 2006 #2 \(\displaystyle \L \begin{array}{rcl} e^{a + bi} & = & e^a (\cos (b) + i\sin (b)) \\ \left| {e^{a + bi} } \right| & = & \left| {e^a (\cos (b) + i\sin (b))} \right| \\ & = & \left| {e^a } \right|\left| {\cos (b) + i\sin (b)} \right| \\ & = & e^a \\ \end{array}\) As a general rule \(\displaystyle \L\left| {e^z } \right| = e^{Re(z)}\)
\(\displaystyle \L \begin{array}{rcl} e^{a + bi} & = & e^a (\cos (b) + i\sin (b)) \\ \left| {e^{a + bi} } \right| & = & \left| {e^a (\cos (b) + i\sin (b))} \right| \\ & = & \left| {e^a } \right|\left| {\cos (b) + i\sin (b)} \right| \\ & = & e^a \\ \end{array}\) As a general rule \(\displaystyle \L\left| {e^z } \right| = e^{Re(z)}\)
L lastlydreaming New member Joined Oct 7, 2006 Messages 10 Oct 10, 2006 #3 So with respect ot all of its arguments, does it become? e^ 2 [(cos 3+ 2pin) + i sin(3+ 2pi n)] where n = 0, +/- , +/- 2.... ? Thanks.
So with respect ot all of its arguments, does it become? e^ 2 [(cos 3+ 2pin) + i sin(3+ 2pi n)] where n = 0, +/- , +/- 2.... ? Thanks.