Find the modulus and principal argument of: 1^° e^(π i/2)…
Find the modulus and principal argument of: $\ 1^\circ\ e^{\pi i/2};\quad2^\circ\ e^{2-3i};\quad3^\circ\ e^{-3-4i};\quad4^\circ\ e^{-i\varphi}\ (|\varphi|<\pi)$.
Explanation of Method
For $z=x+iy$: $\ |e^z|=e^x$ and $\arg(e^z)=y$, reduced modulo $2\pi$ into the principal range $(-\pi,\pi]$.
Part $1^\circ$: $e^{\pi i/2}$
$x=0,\ y=\frac\pi2$. $\ |e^{\pi i/2}|=e^0=1$. Since $\frac\pi2\in(-\pi,\pi]$: $\ \text{Arg}=\frac\pi2$.
Check: $e^{\pi i/2}=\cos\frac\pi2+i\sin\frac\pi2=i$, and indeed $|i|=1,\ \arg i=\frac\pi2$ ✓.
Part $2^\circ$: $e^{2-3i}$
$x=2,\ y=-3$. $\ |e^{2-3i}|=e^2$. Since $-\pi\approx-3.1416<-3\le\pi$, the value $-3$ is already in $(-\pi,\pi]$:
$$ \boxed{|e^{2-3i}|=e^2, \quad \text{Arg}(e^{2-3i})=-3} $$
Part $3^\circ$: $e^{-3-4i}$
$x=-3,\ y=-4$. $\ |e^{-3-4i}|=e^{-3}$. Since $-4<-\pi$, add $2\pi$ to bring it into range:
$$ -4+2\pi\approx2.283\in(-\pi,\pi] $$
$$ \boxed{|e^{-3-4i}|=e^{-3}, \quad \text{Arg}(e^{-3-4i})=2\pi-4} $$
Part $4^\circ$: $e^{-i\varphi}$, $|\varphi|<\pi$
$x=0,\ y=-\varphi$. $\ |e^{-i\varphi}|=1$. Since $|\varphi|<\pi$, we have $-\varphi\in(-\pi,\pi)\subset(-\pi,\pi]$, already principal:
$$ \boxed{|e^{-i\varphi}|=1, \quad \text{Arg}(e^{-i\varphi})=-\varphi} $$
Verification
| Expression | $|e^z|$ | $\text{Arg}(e^z)$ |
|---|---|---|
| $e^{\pi i/2}$ | $1$ | $\frac\pi2$ |
| $e^{2-3i}$ | $e^2$ | $-3$ |
| $e^{-3-4i}$ | $e^{-3}$ | $2\pi-4\approx2.283$ |
| $e^{-i\varphi}$ | $1$ | $-\varphi$ |
Result
The modulus of $e^z$ depends only on $\text{Re}(z)$; the argument is $\text{Im}(z)$ itself, reduced into the principal range by adding or subtracting the appropriate multiple of $2\pi$.