For $z\in \mathbb{C}\setminus [1,\infty )$, the number $w=1-z$ avoids the nonpositive real axis. Define the branch of the complex logarithm byHere the principal argument is selected as a continuous map on that slit domain, so this is a holomorphic function, with derivative $-1/(1-z)$ and $L(1)=0$. For example, locally writing $w=\rho e^{i\varphi }$ gives $d(\log \rho +i\varphi )=dw/w$, verifying the complex derivative. The normalization also makes this branch of the complex logarithm unique: any two continuous choices differ by a constant integer multiple of $2\pi i$ on the connected slit domain, and their values at $w=1$ agree.

To evaluate the corresponding analytic branch of a square root on the unit circle, factorFor $0<\theta <2\pi$, we have $\sin (\theta /2)>0$ and $-\pi /2<\theta /2-\pi /2<\pi /2$. Thus this factorization already uses the selected principal argument, with no hidden change by $2\pi$. It follows thatHalving this expression and applying the exponential function givesSince $\sqrt{2}{e}^{-i\pi /4}=1-i$, the desired branch value isFor instance, $\theta =\pi$ gives $\sqrt{2}>0$, a quick check that the sign agrees with the normalization.

List the roots with their multiplicities. By the factor theorem, $Q(z)=a{\prod }_{k=1}^n(z-{\alpha }_k)$ with $a\ne 0$. The product rule, followed by division where $Q(z)\ne 0$, gives its logarithmic derivative:Let $Q^{\prime }(\beta )=0$. If $Q(\beta )=0$ too, then $\beta$ is already one of the roots, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all roots have positive imaginary parts but $\mathrm{Im}\beta \le 0$, thenfor every $k$. Their sum cannot be zero. Thus $\mathrm{Im}\beta >0$.

For the unit disk conclusion, the same sum gives an especially transparent convex combination. Taking complex conjugates and setting $w_k=|\beta -{\alpha }_k{|}^{-2}>0$ givesTherefore, if every $|{\alpha }_k|<1$,The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the Gauss-Lucas theorem: every critical point of a nonconstant polynomial lies in the convex hull of its roots. The common-root case, handled before division, is essential when multiplicities occur.

Put $S={\sum }_{j=0}^{n-1}|a_j|$ and choose $R>\max (1,2S/|a_n|)$. For $|z|>R$, the triangle inequality givesThe reverse triangle inequality therefore yieldsIf $P$ had no root, the function $h=1/P$ would be an entire function. The bound just proved makes $h(z)\to 0$ as $|z|\to \infty$. On the remaining closed disk, continuity and the extreme value theorem make $h$ bounded. Thus $h$ is a bounded entire function, so Liouville's theorem makes it constant. Its limit at infinity forces that constant to be zero, impossible for a reciprocal.

Consequently $P$ has a root in $\mathbb{C}$. This proves the existence part of the fundamental theorem of algebra. The invoked Liouville theorem follows from the Cauchy integral formula: the Cauchy estimate $|h^{\prime }(w)|\le M/r$ on arbitrarily large circles gives $h^{\prime }=0$ when $h$ is bounded by $M$.

One standard form of the Cauchy integral formula is this: if $f$ is holomorphic on an open set containing a contour that is a simple closed curve $\Gamma$ and its interior, and $\Gamma$ has positive orientation of a contour, then for every point $w$ inside it,The hypothesis includes the contour itself; singularities on the contour are excluded. In particular, the formula applies on a circle whose closed disk is contained in the domain of holomorphy.

Define $g(p)=f(p)-f(-p)$ on the sphere. It is continuous and satisfies $g(-p)=-g(p)$. If $g(p)=0$ at a chosen point, we are done. Otherwise choose a unit vector $q$ perpendicular to $p$ and follow the half great circleThe Pythagorean theorem shows that $\gamma (t)$ lies on the sphere, and its endpoints are $p,-p$. The continuous function $g\circ \gamma$ has opposite signs at the endpoints. The Intermediate value theorem therefore gives a zero $t_{\ast }$, soThis is a scalar equal-value assertion for an antipodal pair; it uses only a path and the Intermediate value theorem.

Suppose a path-connected space had a separation $X=U\cup V$. Choose $u\in U$, $v\in V$, and a path $\gamma$ from $u$ to $v$. The inverse images ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$ are disjoint open sets in the subspace topology of $[0,1]$, cover it, and are nonempty because they contain its opposite endpoints. This contradicts the connectedness of an interval.

Therefore every path-connected space is connected. Notice that it is the subspace topology on $[0,1]$ that is used in the inverse-image argument.

We can arrange that neither the supremum nor the infimum is attained. By part (d), choose a continuous function $f$ on $X$ with unbounded range, and put $u=|f|$. Choose $x_n\in X$ so that $t_n=u(x_n)$ increases strictly to infinity. Construct a continuous function $h:[0,\infty )\to \mathbb{R}$ withkeep it constant up to $t_1$ and interpolate linearly on each $[t_n,t_{n+1}]$. Since $t_n\to \infty$, the joining points have no finite accumulation, so this defines a continuous function everywhere on its domain.

Now setBy function composition, $g$ is continuous, and $-1<g(x)<1$ for every $x$. Along the even and odd chosen points,Thus ${\sup }_{X}g=1$ and ${\inf }_{X}g=-1$, and neither is a value of $g$. This proves the bounded continuous witness with neither extremum attained construction, including for subsets that are bounded but not closed.

By the Heine-Borel theorem, a noncompact subset $X\subseteq \mathbb{R}$ is either unbounded or not closed. If $X$ is unbounded, the restriction $f(x)=x$ is a continuous function with unbounded range.

Otherwise $X$ is bounded and not closed. Choose $a\in \overline{X}\setminus X$, and defineIts denominator is nonzero on $X$, so it is continuous. Since $a$ is in the closure, for every $n$ there is $x_n\in X$ with $0<|x_n-a|<1/n$. Hence $f(x_n)>n$, proving that this continuous function is unbounded. These two cases establish unbounded continuous witnesses to noncompactness of a real subspace.