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A real matrix is orthogonal when $Q^{T}Q=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda |=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian inner product gives $(1-\overline{\lambda }\mu )v^{†}w=0$; distinct unit-modulus eigenvalues therefore have orthogonal eigenvectors.

The nonreal eigenvalues of a real $3\times 3$ matrix occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real eigenvalue is $-1$. If $x\cdot n=0$, then $(Qx)\cdot (Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose determinant is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal basis adapted to $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

The comparison test says that if $0\le b_n\le c_n$ eventually and $\sum c_n$ converges, then $\sum b_n$ converges. If $\sum a_n{z}_0^n$ converges, its terms are bounded: $|a_n{z}_0^n|\le M$. For $|z_1|<|z_0|$,so comparison with a geometric series proves absolute convergence.

The radius $R$ is the number for which the power series converges absolutely for $|z|<R$ and diverges for $|z|>R$. Given $r_i<R_i$, both $\sum |a_n|r_1^n$ and $\sum |b_n|r_2^n$ converge, so their terms are bounded, say $|a_n|r_1^n\le M$. ThenThus $R\ge r_1{r}_2$; letting $r_i\uparrow R_i$ gives $R\ge R_1{R}_2$.

If $f$ is differentiable at $a$ and $g$ at $f(a)$, then $g\circ f$ is differentiable at $a$ andWrite $g(f(a)+u)-g(f(a))=u\{g^{\prime }(f(a))+\epsilon (u)\}$, where $\epsilon (u)\to 0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the limit proves the formula.