Solution. Write \(w=x+iy\), so that \(e^w=e^x(\cos y+i\sin y)\), giving \(\left |e^w\right |=e^x\) and \(\arg e^w=y\). Matching modulus and argument turns each equation into two real ones, and because the argument is fixed only up to \(2n\pi \) every equation has infinitely many solutions.

(i)

\(\left |i\right |=1\), \(\arg i=\frac {\pi }{2}\), so \(x=0\) and \(y=\frac {\pi }{2}+2n\pi \): \[w=\left (\frac {\pi }{2}+2n\pi \right )i,\qquad n\in \mathbb {Z}.\]

(ii)

\(\left |1+i\right |=\sqrt 2\), \(\arg (1+i)=\frac {\pi }{4}\), so \(x=\frac {1}{2}\ln 2\) and \(y=\frac {\pi }{4}+2n\pi \): \[w=\frac {1}{2}\ln 2+\left (\frac {\pi }{4}+2n\pi \right )i,\qquad n\in \mathbb {Z}.\]

(iii)

\(\left |-2\right |=2\), \(\arg (-2)=\pi \), so \(x=\ln 2\) and \(y=(2n+1)\pi \): \[w=\ln 2+(2n+1)\pi i,\qquad n\in \mathbb {Z}.\]

Part (iii) is worth pausing on: \(e^w\) takes a negative value, although \(e^x\) never does for real \(x\). The complex exponential is onto \(\mathbb {C}\setminus \{0\}\); only \(0\) is missed.

Proof. We compute that \(\left |e^z\right | = \left |e^{x + iy}\right | = e^x\) and \(Arg(e^z) = y + 2\pi n, \quad n\in \mathbb {Z}\), putting them in the definition, we have \begin {align*} \log (e^z) & = \ln \left |e^z\right | + iArg(e^z)\\ & = \ln e^x + i (y + 2n\pi )\\ & = x + iy + 2n\pi i\\ & = z + 2n\pi i, \quad n\in \mathbb {Z} \end {align*}

The logarithmic function is a multi- valued. We restrict the argument in the logarithmic function and deduce a single valued function. □

Solution. The principal value takes the principal argument, \(-\pi <\operatorname {Arg}z\leq \pi \), so \(\operatorname {Ln}z=\ln \left |z\right |+i\operatorname {Arg}z\). Only the argument needs care.

(i)

\(\operatorname {Ln}i=\ln 1+\frac {\pi }{2}i=\frac {\pi }{2}i\).

(ii)

\(\operatorname {Ln}(1+i)=\frac {1}{2}\ln 2+\frac {\pi }{4}i \approx 0.3466+0.7854\,i\).

(iii)

\(\operatorname {Ln}(-2)=\ln 2+\pi i\approx 0.6931+3.1416\,i\).

Compare with the previous example: there the answers carried a free \(2n\pi \); here the principal branch selects one of them. The price of single-valuedness is the cut along the negative real axis, across which \(\operatorname {Arg}\) jumps by \(2\pi \).

Proof. Two things must be checked: that the restricted exponential is one-to-one, so that an inverse exists at all, and that its inverse is the principal logarithm.

Injective on the strip

Suppose \(e^{z_1}=e^{z_2}\) with both \(z_j=x_j+iy_j\) in the strip \(-\pi <y\leq \pi \). Then \(e^{z_1-z_2}=1\), so \(e^{x_1-x_2}=1\) and \(y_1-y_2=2n\pi \) for some integer \(n\). The first gives \(x_1=x_2\), since the real exponential is injective. For the second, both \(y_1\) and \(y_2\) lie in an interval of length \(2\pi \), so \(\left |y_1-y_2\right |<2\pi \), forcing \(n=0\) and \(y_1=y_2\). Hence \(z_1=z_2\).

The inverse is \(\operatorname {Ln}\)

Given \(w\neq 0\), write \(w=\left |w\right |e^{i\operatorname {Arg}w}\) with \(-\pi <\operatorname {Arg}w\leq \pi \), and set \(z=\ln \left |w\right |+i\operatorname {Arg}w\). Then \(z\) lies in the strip and \(e^{z}=w\), so the map is onto \(\mathbb {C}\setminus \{0\}\) and its inverse sends \(w\) to \(\operatorname {Ln}w\).

The value \(0\) is omitted because \(\left |e^{z}\right |=e^{x}>0\) always, which is why the logarithm has no value there. □

Proof. Note that \(f_1\) is defined on the domain \(\left |z\right | > 0,\, -\pi \leq \operatorname {arg}(z) \leq \pi \quad \cdots \quad (*)\). In polar co ordinates, \(f_1 = \log _e r + \theta i\). So we can set \(u(r,\theta ) = \log _e r\quad ,\quad v(r,\theta ) = \theta \).
Note that, the Cauchy-Riemann equations in polar coordinates are given by \[\frac {\partial u}{\partial r} = \frac {1}{r}\,\frac {\partial v}{\partial \theta }\qquad ,\qquad \frac {\partial v}{\partial r} = \frac {-1}{r}\,\frac {\partial u}{\partial \theta }\] Now since \(u(r,\theta ) = \log _er\) and \(v(r,\theta ) = \theta \) we have \begin {align*} \frac {\partial u}{\partial r} & = \frac {1}{r}\qquad ,\qquad \frac {\partial v}{\partial \theta } = 1\\\\ \frac {\partial v}{\partial r} & = 0\qquad ,\qquad \frac {\partial u}{\partial \theta } = 0 \end {align*}

So we see that \(f_1\) satisfies the Cauchy-Riemann equations in polar form. Also \(u, v\) and the first partial derivatives of \(u\) and \(v\) are continuous at all points in the domain \((*)\).
It follows that \(f_1\) is analytic in this domain. We now compute the derivative of \(f_1\). Note that \(\, z = re^{i\theta }\). So \[\frac {\partial f_1}{\partial r}= \frac {\partial f_1}{\partial z}\cdot \frac {\partial z}{\partial r}\implies \frac {\partial f_1}{\partial r}=\frac {\partial f_1}{\partial z}\cdot e^{i\theta }\implies \frac {\partial f_1}{\partial z} = \frac {e^{i\theta }\,\partial f_1}{\partial r}\]

Recall that \(f'_1(r) = \frac {\partial u}{\partial r} + i\frac {\partial v}{\partial r}.\,\) This means that \[\frac {\partial f_1}{\partial z} = e^{-i\theta }\Big (\frac {1}{r} + 0\Big ) = \frac {1}{re^{i\theta }} = \frac {1}{z}\] □