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A complex number can encode both the wave amplitude and phase of a harmonic disturbance: $a(x,t)=\mathrm{Re}\{\hat{a}{e}^{i(kx-\omega t)}\}$. Its modulus is the real oscillation amplitude and its complex argument fixes the wave phase.

For every integer $n$ and real $\theta$,It converts powers of a complex number in polar form into multiplication of its complex argument.

For nonzero complex numbers $x,y$, equality $|x+y|=|x|+|y|$ holds exactly when $x/y$ is a positive real number, equivalently when the two numbers have the same complex argument modulo $2\pi$.

Similarly,Thus, for $0<\theta <\pi$,At $\theta =0$, $1-z=0$, whose complex argument is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin (\theta /2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

Let $u(t)=\gamma (t)/|\gamma (t)|$, a continuous map into the unit circle. By uniform continuity, partition $[0,1]$ so finely that on each interval beginning at $t_j$, $|u(t)/u(t_j)-1|<1$. This ratio lies in the right half-plane and has a continuous complex argument with value zero at $t_j$. Starting from ${\theta }_0$, add these local arguments successively, matching endpoint values. This constructs a continuous real lift with $u(t)=e^{i\theta (t)}$.

Any two such lifts with the same initial value differ by a continuous function taking values in $2\pi \mathbb{Z}$, and hence coincide. Since the curve closes, $e^{i(\theta (1)-\theta (0))}=1$. Define its winding number byChanging the chosen initial argument adds a constant multiple of $2\pi$ to the whole lift, so the winding number is independent of that choice. No differentiability of the curve is required.

The continuous closed curve $q(t)={\gamma }_2(t)/{\gamma }_1(t)$ satisfies $|q(t)-1|<1$. It lies in the right half-plane, which admits a single continuous complex argument in $(-\pi /2,\pi /2)$. Since $q$ closes, this argument has equal endpoint values, giving $w(q)=0$. Additivity under multiplication then yieldsThe strict inequality ensures that the quotient stays in a region avoiding zero and admitting a single argument branch.

For $z=x+iy$,Matching its modulus and complex argument with $2e^{i\pi /6}$ gives

Write $z=r{e}^{i\theta }$ with $-\pi <\theta \le \pi$. Its principal sixth root is $r^{1/6}{e}^{i\theta /6}$, whose complex argument lies in $(-\pi /6,\pi /6]$. Equality with $2e^{i\pi /6}$ forces $r=64$ and $\theta =\pi$, so $z=-64$.

All values of the complex logarithm areBecause $x>0$, one may take the complex argument $\arg z=\arctan (y/x)$. Consequently

By the equality condition in the complex triangle inequality,for nonzero $x,y$ exactly when $x/y$ is a positive real number. Having the same complex argument is reflexive, symmetric, and transitive, so this is an equivalence relation. Its classes are the open rays from the origin.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

The principal cube root is $z^{1/3}=\exp (\mathrm{Log}z/3)$ using the principal complex logarithm. In the right half-plane, its complex argument lies between $-\pi /6$ and $\pi /6$. Consequently the roots of $r^3=-z$ consist of one root with negative real part, $-z^{1/3}$, and two with positive real part, $z^{1/3}{e}^{\pm i\pi /3}$.