Recipe

Evaluate the principal logarithm \operatorname{Log}z

Use when: you need \operatorname{Log}z for a specific z\neq0, or you're comparing \operatorname{Log}(z^n) against n\operatorname{Log}z

Goal: the single principal value \ln|z|+i\operatorname{Arg}z, with \operatorname{Arg}z correctly picked from (-\pi,\pi]

Taught in The logarithm and its branches

1. 1

   Compute the modulus |z| the same way as always, from z's real and imaginary parts.

   Example

   For z=1-i: |z|=\sqrt{1^2+(-1)^2}=\sqrt2.

   Why: \ln|z| is the real part of \operatorname{Log}z, and it's undefined only at z=0, exactly where |z|=0.

2. 2

   Find the reference angle from the ratio of z's imaginary and real parts, then adjust it by quadrant so the result lands inside (-\pi,\pi].

   Example

   For z=1-i: \operatorname{Re}(z)=1>0, so the reference angle is \arctan(1)=\pi/4, and since \operatorname{Im}(z)=-1<0, \operatorname{Arg}(1-i)=-\pi/4.

   • \operatorname{Re}(z)>0 — \operatorname{Arg}z is the reference angle itself, signed to match \operatorname{Im}(z).
   • \operatorname{Re}(z)<0 and \operatorname{Im}(z)\neq0 — \operatorname{Arg}z is \pi minus the reference angle when \operatorname{Im}(z)>0, or the reference angle minus \pi when \operatorname{Im}(z)<0.
   • z sits exactly on the negative real axis (\operatorname{Re}(z)<0, \operatorname{Im}(z)=0) — \operatorname{Arg}z, and therefore \operatorname{Log}z, is undefined; z sits on the branch cut, outside \operatorname{Log}'s domain.

   Why: the reference angle alone is always a positive acute angle; it's the quadrant, not the reference angle, that decides the correct sign and offset.

3. 3

   Plug the modulus and principal argument from Steps 1 and 2 into the definition.

   Example

   \operatorname{Log}(1-i)=\ln\sqrt2-i\frac{\pi}{4}=\frac12\ln2-\frac{\pi}{4}i.

   Why: this is exactly the definition being applied, once both pieces it needs are on hand.

4. 4

   Compute n\operatorname{Arg}z and check whether it already sits inside (-\pi,\pi]; that comparison, not the algebra, is what decides whether the two sides agree.

   Example

   For z=-1-i: \operatorname{Arg}z=-3\pi/4, so n\operatorname{Arg}z=2\cdot(-3\pi/4)=-3\pi/2, which is not in (-\pi,\pi] (it's past -\pi). Indeed z^2=2i has \operatorname{Arg}(z^2)=\pi/2=-3\pi/2+2\pi, so \operatorname{Log}(z^2)-2\operatorname{Log}(z)=2\pi i, one full period, never anything messier.

   • n\operatorname{Arg}z\in(-\pi,\pi] — it already equals \operatorname{Arg}(z^n), so \operatorname{Log}(z^n)=n\operatorname{Log}z holds exactly.
   • n\operatorname{Arg}z falls outside (-\pi,\pi] — \operatorname{Arg}(z^n) is n\operatorname{Arg}z shifted by a multiple of 2\pi back into range, so \operatorname{Log}(z^n) and n\operatorname{Log}z differ by that same multiple of 2\pi i.

   Why: \operatorname{Arg} is only single-valued because it's been squeezed into one interval by hand, and multiplying an angle by n can easily push the true angle back out of that interval; that's the entire mechanism behind the log law failing.