If I use the Euler notation for a complex number, then a complex number $z$ can the following exponential form:
$$e^z=e^{x+iy}=e^x e^{iy}=e^x(\cos y+i\sin y)$$
So $e^x$ and $y$ represent the module and the angle of polar form of $z$, respectively?
Asked
Active
Viewed 185 times
3
-
3Yes, exactly... – TheGeekGreek Apr 20 '17 at 21:03
-
Ok, but what if I multiply $e^z$ by, say, $\alpha=\frac{1}{3}(cos(-\frac{\pi}{3})+isen(-\frac{\pi}{3}))$, i.e., a number in polar form, how do I get the polar form again? Will it be just $\alpha z=\frac{1}{3}e^x(cos(y-\frac{\pi}{3})+isen(y-\frac{\pi}{3}))?$. I need to get the cartesian coordinates of this and it just doesn't seems right to me. – AnalyticHarmony Apr 20 '17 at 21:09
-
2@A.Mello: Yeap! It is right! – Arnaldo Apr 20 '17 at 21:11
-
1@A.Mello $e^{i \pi/2}$ is the Euler form of $z=i$ and not of $z=i\pi/2$. – AnyAD Apr 20 '17 at 21:27
-
@Any So if $z=x+iy=|z|(cos \theta + i sen \theta)=|z|e^{i \theta}$, what's the point in having $e^x$ as module of $z = x+iy$? – AnalyticHarmony Apr 20 '17 at 21:56
-
1@A.Mello The modulus of $0+1×i $ is $e^0=1$. – AnyAD Apr 20 '17 at 23:24
-
1That is, $z $ is represented by $re^{i \theta}$ if $r=|z|$ and $\theta = arg (z)$. – AnyAD Apr 20 '17 at 23:26
-
1But you may want to ignore these comments, if I have misundrstood your question – AnyAD Apr 20 '17 at 23:32