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Expand the sine of the complex number using hyperbolic functions:Its squared complex modulus isTaking nonnegative square roots proves $\boxed{|\sin z|\ge |\sin x|}$. Equality holds exactly when $y=0$.

Use these standard facts about a power series: it is absolutely convergent at every point strictly inside its radius of convergence and divergent at every point strictly outside it. Also, the triangle inequality implies that a sum or difference of two absolutely convergent series is absolutely convergent.

Let $T$ be the radius of convergence of the coefficientwise sum. For $|z|<\min (R,S)$,Hence $T\ge \min (R,S)$.

Suppose $R<S$. If $T>R$, choose a finite $\rho$ with $R<\rho <\min (T,S)$ and choose $z$ of complex modulus $\rho$. The sum series and the $b_n$ series both converge absolutely there, socontradicting $\rho >R$. Thus $T=R$. Interchanging the series covers $S<R$, proving the sum of power series with unequal radii of convergence rule:The argument also covers a zero radius or an infinite larger radius. When the two radii are equal, cancellation can enlarge the radius, so the inequality of the radii matters.

The polar form of a complex number and the roots of unity givesince $512=2^9$. Using the complex conjugate to compute the squared complex modulus,The residue $3j-k$ runs through every class modulo $9$, already with $j=0$. The cosine is even, so the nine angles give five distinct values, represented by $r=0,1,2,3,4$. They are distinct because cosine is strictly decreasing on $[0,\pi ]$ and these representative angles lie in that interval. HenceEvery listed value is attained by choosing $z_1=1$ and $z_2=2e^{2\pi ir/9}$.

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives

Away from the negative real axis, the complex conjugate has principal complex argument $-\theta$. ConsequentlyTaking the complex modulus gives $\cosh \theta =1$, hence $\theta =0$. The remaining equation is $e^{i\log r}=i$, soOn the negative real axis, both $z$ and $\overline{z}$ have principal complex argument $\pi$, so the left-hand side has modulus $2e^{-\pi }$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline{z}$, not the complex conjugate of $z^i$.

The principal complex exponentiation gives $z^i=e^{-\theta }{e}^{i\log r}$. Equality to one requires its complex modulus to be one, hence $\theta =0$, and then its complex argument requires $\log r\in 2\pi \mathbb{Z}$. Thus all solutions are positive real, namelySubstitution verifies every value, and the modulus condition excludes every nonreal value.

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

Write $z=1+iy$ with $y\in \mathbb{R}$. In the complex plane, this is the vertical line through $1$. The complex exponential gives $e^z=e(\cos y+i\sin y)$, so its complex modulus is always $e$. Conversely, every point of that modulus has an complex argument $y$ and is obtained in this way. The image is the entire circle centered at zero with radius $e$, traversed repeatedly as $y$ varies; it is not a disk.

For $z=x+iy$, the complex exponential has complex modulus $e^x$. Equality to the positive real number $e$ therefore forces $x=1$ and $\cos y+i\sin y=1$. The latter holds precisely when $y$ is an integer multiple of $2\pi$. Hence all solutions areSubstitution verifies every listed value; the modulus and angle arguments exclude any others.

Put $z=x+iy$. Comparing imaginary parts gives $-y=1$, so $y=-1$. Comparing real parts then gives $\sqrt{x^2+1}=x$. The left side is positive, so squaring is legitimate only with $x\ge 0$, but even then it would give $x^2+1=x^2$. This contradiction proves there is no solution. Equivalently, with nonzero imaginary part the complex modulus is strictly greater than the real part.

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

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.

Write $z=x+iy$, with $y\ne 0$ because the triangle is nondegenerate. Let $w=u+iv$ be its circumcenter. Equality of the complex modulus values $|w|=|w-1|$ gives $u=1/2$. Equality of $|w|=|w-z|$, after expanding the squares, givesThus the real part and imaginary part determineUsing the complex conjugate $\bar{z}=x-iy$, we also haveThereforeThe nonzero denominator is exactly the nondegeneracy condition. If $z$ is real, the three distinct collinear points do not have a finite circumcenter.

The logarithm's argument is $\sqrt{3}/2+i/2=e^{i\pi /6}$, with complex modulus one. The principal complex logarithm is therefore $\ln 1+i\pi /6=i\pi /6$, givingIf $\ln z$ denotes all logarithm values rather than the principal branch, they are $i(\pi /6+2\pi k)$ for $k\in \mathbb{Z}$: the real part remains zero and the imaginary part changes by integer multiples of $2\pi$. The principal value is the conventional single-valued answer.

The point has positive real part and negative imaginary part. Its complex modulus is $\sqrt{1+3}=2$, and its principal complex argument is $-\pi /3$. Hence

Write $z=x+iy$. The complex modulus condition is equivalent to $(x-1{)}^2+y^2=4$. Thus the locus is a circle of radius two centred at $(1,0)$ in the complex plane. Its extreme points are $(-1,0)$, $(3,0)$, and $(1,\pm 2)$.

Write a square root as $u+iv$, with $u,v$ real. Equating the real part and imaginary part of its square gives $u^2-v^2=-1$ and $2uv=1$. Taking complex modulus also gives $u^2+v^2=\sqrt{2}$. ThereforeSince $uv>0$, the signs agree. The two roots areThe $\pm$ changes both signs together, rather than independently.

Using the exponential definition of hyperbolic cosine, multiply $e^z+e^{-z}=-2$ by the nonzero quantity $e^z$. This yields $(e^z+1{)}^2=0$, hence $e^z=-1$.

If $z=x+iy$, its complex modulus requires $e^x=1$, so $x=0$; its phase then requires $y=(2k+1)\pi$. Thus all complex solutions areThere are no real solutions, since real $\cosh x\ge 1$.

The original PDF asks for cube roots of $8$. Their complex modulus is $2$, and their arguments are $2k\pi /3$, $k=0,1,2$. Converting to Cartesian form givesCubing any one gives the positive real number $8$.

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.