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A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

A scalar $\lambda$ is an eigenvalue of $A$ if and only if there is a nonzero vector $v$ with $Av=\lambda v$. By the preceding kernel criterion,The last condition is ${\chi }_A(\lambda )=0$, exactly the definition of a polynomial root. For a real matrix, this proves the assertion for real eigenvalues; the same proof over the complex numbers gives the usual complex version.

Let $A$ be a skew-symmetric matrix. For every real vector $x$, the real scalar $x^{T}Ax$ equals its own transpose, butHence $x^{T}Ax=0$. If $Av=\lambda v$ is a real eigenvector equation with $v\ne 0$, it follows thatso $\lambda =0$. If $A$ is a diagonalisable matrix over $\mathbb{R}$, it has a real basis of eigenvectors, all with eigenvalue zero. Thus $A$ vanishes on a basis and $\boxed{A=0_n}$.

The field matters: a nonzero real skew-symmetric matrix can be a diagonalisable matrix over the complex numbers, for example $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ with eigenvalues $i,-i$.

The exponential function definitions of the hyperbolic functions and Euler's formula giveDividing these identities wherever $\cos x\ne 0$ gives $\boxed{\tanh (ix)=i\tan x}$. The two undivided identities hold for every complex number $x$; the quotient identity excludes its poles.

Adding and subtracting the exponential function power series cancels the appropriate odd or even terms:Both series converge absolutely for every complex number $x$, as follows from the ratio test.

For a matrix proportional to its transpose, applying the transpose twice gives $A^T=5A$ and therefore $A=25A$. Over the real numbers or complex numbers, $24A=0$ implies

Since $(1-i{)}^2=-2i$, division in the complex numbers gives

The complex number $-1+i$ has complex modulus $\sqrt{2}$ and argument $3\pi /4$. A cube root has modulus $(\sqrt{2}{)}^{1/3}=2^{1/6}$, and its argument is one third of $3\pi /4+2k\pi$. Hence the three distinct roots areTheir displayed arguments are $\pi /4$, $11\pi /12$, and $19\pi /12$; the last is equivalently $-5\pi /12$ in the principal range.

The factor $1+i$ has modulus $\sqrt{2}$ and principal argument $\pi /4$. Multiplication by $e^{i\pi /6}$ adds $\pi /6$ to its complex argument, while the real factor $1/2$ halves its modulus. HenceTaking the principal logarithm for $w$ givesThis complex number lies in the second quadrant. Its modulus follows from the sum of squared components, and its principal argument follows by the second-quadrant correction:All arguments of this fixed $w$ add integer multiples of $2\pi$ to the displayed one.

If $\ln z$ is interpreted as the multivalued complex logarithm, there is instead a family $w_k=-a+i{b}_k$, where $a=\frac{1}{2}\ln 2>0$ and $b_k=5\pi /12+2\pi k$. Then $|w_k|=\sqrt{a^2+b_k^2}$, and the principal argument is $\pi -\arctan (b_k/a)$ for $b_k>0$, or $-\pi -\arctan (b_k/a)$ for $b_k<0$. This records the logarithm-branch choice that the paper leaves implicit.

Over the complex numbers, the polynomial factorization $z^2+1=(z-i)(z+i)$ gives the partial fraction decompositionMultiplication by $(z-i)(z+i)$ and comparison of polynomial coefficients give $A+B=1$ and $i(A-B)=1$. Hence

The literal expression in the original PDF is undefined. Indeed, hyperbolic tangent satisfies $\tanh w=\sinh w/\cosh w$, andAlready the $n=2$ summand contains $\tanh (i\pi /2)$, a pole. The shifted term has the same issue when $n\equiv 3(\mathrm{mod}4)$. Thus this is not a finite sum of defined complex numbers, and it has no literal real or imaginary part. Cancelling two undefined terms is not ordinary arithmetic.

There is nevertheless a natural interpretation by meromorphic telescoping cancellation. Insert the same small positive real shift $\epsilon$ into every hyperbolic tangent argument. Each term is then finite, and the telescoping sum isSince hyperbolic tangent has period $i\pi$, its first argument can be reduced to $\epsilon +i\pi /4$. Taking the limit givesThis is a common-shift continuation of the sum, plausibly the intended telescoping answer. It requires that additional convention; it is not the value of the expression as printed.

The first term is $ar$, so the finite geometric series is $S=a(r+r^2+\cdots +r^{100})$. Subtracting $rS$ cancels the interior terms and gives $(1-r)S=ar(1-r^{100})$. ThereforeThe rational expression holds for every real $a$ and every real $r\ne 1$, including $r=0$; the same algebra works for complex numbers. No restriction $|r|<1$ is needed for a finite geometric series. The exceptional case follows by adding one hundred equal terms, or by taking the limit as $r\to 1$.

The given complex numbers have positive real parts, so $\arg z_1=\arctan A$ and $\arg z_2=\arctan B$. Their product isBy the assumed complex argument ranges, this product also has positive real part, hence $1-AB>0$. Its complex argument is consequently $\arctan [(A+B)/(1-AB)]$, with no extra multiple of $\pi$. Since multiplication adds complex arguments,This proves the arctangent addition formula under the stated branch restriction. For arbitrary real $A,B$, the principal inverse tangent formula can require adding or subtracting $\pi$ when $AB>1$.

Multiplying the complex numbers gives $b\bar{z}+\bar{b}z=2(b_{1}x+b_{2}y)$. Hence the real linear equation is

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.

For the translation, expand the products of complex numbers and collect the terms in $w\bar{w}$, $\bar{w}$, $w$ and the constant:It follows thatThe last coefficient is real because $b\bar{\alpha }+\bar{b}\alpha =2\mathrm{Re}(\bar{b}\alpha )$.

A square matrix is invertible precisely when its determinant is nonzero. Equivalent conditions are full rank, a trivial kernel, and zero not being an eigenvalue.

For a lower triangular matrix, the first row has only the possibly nonzero entry $A_{11}$. A Laplace expansion along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yieldsThis proof includes singular triangular matrices; it makes no division by diagonal entries.

For the diagonal example, put $N=n(n+1)/2$. Its determinant is $b^{1+2+\cdots +n}=b^N$, and invertibility requires $b\ne 0$. Write the complex number as $b=r{e}^{i\theta }$ with $r>0$. A nonzero purely imaginary determinant requires $\cos (N\theta )=0$. ThusEquivalently $b^N\in i\mathbb{R}\setminus \{0\}$; the angle is understood modulo $2\pi$.

For skew-Hermitian matrices, reversing products under the conjugate transpose gives $(CD{)}^{†}=D^{†}{C}^{†}=DC$. ConsequentlyThe latter condition is vanishing of the anticommutator. A universal choice for the last request isIndeed $(CD-DC{)}^{†}=DC-CD=-(CD-DC)$, so the commutator of two skew-Hermitian matrices is skew-Hermitian.

Use $z\overline{z}=|z{|}^2$ and $\overline{B}z+B\overline{z}=2\mathrm{Re}(\overline{B}z)$. When $A\ne 0$, completing the square givesThus, with $\Delta =|B{|}^2-AC$,If $\Delta >0$, this is a circle with centre $-B/A$ and radius $\sqrt{\Delta }/|A|$. If $\Delta =0$, it is the single point $-B/A$, sometimes called a degenerate circle of radius zero. If $\Delta <0$, it is the empty set, because a squared complex modulus cannot be negative.

When $A=0$ and $B\ne 0$, write $B=u+iv$ and $z=x+iy$. The remaining equation iswhich is a straight line with normal vector $(u,v)$. It passes through zero exactly when $C=0$. Finally, when $A=B=0$, the equation has no solutions if $C\ne 0$ and every complex number is a solution if $C=0$.

The complete classification of a real complex-circle equation therefore includes two degeneracies in addition to ordinary circles, straight lines and the empty set:For the three advertised alternatives to be exhaustive, one must exclude the identically zero equation and either exclude or explicitly allow radius-zero circles. This is a qualification of the unrestricted coefficients, not a reason to discard the genuine circle and straight line cases used next.

Use phasors with time dependence $e^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is the complex number relating the sinusoidal voltage and electric current: $|Z|$ is their amplitude ratio and $\arg Z$ is the voltage phase relative to the electric current. The component laws giveThus the resistor has zero relative phase, the capacitor voltage lags its electric current by $\pi /2$, and the inductor voltage leads its electric current by $\pi /2$. The ideal capacitor and inductor store and return energy; the real electrical impedance of the resistor describes dissipation.

Use phasors with time dependence $e^{i\omega t}$ and electrical impedance $Z=\tilde{V}/\tilde{I}$. The element laws $V=IR$, $I=CdV/dt$ and $V=LdI/dt$ giveA complex number records both amplitude and phase, turning the linear differential equations of a sinusoidally driven electrical circuit into algebraic equations. A resistor has in-phase voltage and electric current and dissipates average power. For an ideal capacitor, the electric current leads the voltage by $90^{\circ }$; for an ideal inductor, it lags by $90^{\circ }$. Their imaginary electrical impedances express reversible energy storage and return, with zero net average power over a period. Using $e^{-i\omega t}$ would conjugate the displayed reactive signs without changing predictions.

In the illustrated Maxwell bridge, the left and right vertices are the supply terminals and the upper and lower vertices are detector terminals. Insert a sensitive AC null detector between the latter and drive the former with a sinusoidal source. The upper-left arm is $Z_x=R_x+i\omega L_x$, the upper-right arm is $R_3$, the lower-left arm is $R_2$, and the lower-right arm is the parallel combinationAt a null no detector electric current flows, so the two voltage dividers giveMultiplying by the parallel-arm electrical admittance yieldsEquating real and imaginary parts therefore givesTwo independent adjustments are needed to annul the in-phase and quadrature detector signals. For example, change $R_2$ to set $L_x$ and $R_1$ to set $R_x$, iterating until the detector reads zero. Alternatively, with fixed $R_2,R_3$, independently vary $R_1$ and $C$.

The Maxwell bridge compares an unknown lossy inductor with a stable standard capacitor, avoiding the need for an accurately known reference inductor. The ideal balance equations contain no frequency, and a null measurement need not rely on absolute detector calibration. In practice choose a suitable frequency below the coil's self-resonance, where the series $R_x,L_x$ model and the standard capacitor remain accurate.

If the initial setting is far from balance, the bridge as drawn needs adjustable components and a detector, not merely a more precise readout. Fit resistance boxes or switched ranges for $R_1$ and a ratio arm, and a selectable standard $C$. Choose ranges aroundThen approach the null with coarse adjustments followed by finer ones. A large required $R_1$ can be brought into range by increasing $C$ and decreasing $R_2{R}_3$ proportionally. Start with reduced excitation or reduced detector sensitivity to avoid overload, then increase sensitivity near the null. Changing frequency alone cannot repair an impossible component range, since the ideal balance conditions are frequency-independent. Stray capacitance and lead electrical resistance set practical accuracy limits and may require shielding and short connections.