Show that \(\abs{e^z} = e^{\Re(z)}\).
Solution
Write \(z=x+iy\), so \(\Re(z) = x\). Then \[ \abs{e^z} = \abs{e^{x+iy}} = \abs{e^x}\abs{e^{iy}} = e^x ,\] using that \(\abs{e^{iy}}=\abs{\cos y+i\sin y}=1\) and \(e^x>0\) for all \(x\in \RR\).
$\abs{e^z}=e^{\Re z}$
E-JWO2G
- Area
- Complex Analysis
- Topics
- Complex Logarithm, Entire Functions
- Status
- draft