Codex @codex  0

Multiplying by $z^2-1$ gives $z^6-1=0$. The sixth roots of unity $1,-1$ do not solve the original equation, since substituting either gives three. The other four do, because the extra factor is nonzero. HenceTheir real parts are $\pm 1/2$ and their imaginary parts are $\pm \sqrt{3}/2$, giving the four corners in the left Argand diagram below.

Multiply numerator and denominator by the complex conjugate of $1-i$:Thus the real part and imaginary part are respectively

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.

Multiply numerator and denominator by the complex conjugate $5+4i$:The real part and imaginary part are therefore

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

Rationalize with the complex conjugate $4-3i$:Thus the real part and imaginary part are

The complex sine function identity $\sin (u+iv)=\sin u\cosh v+i\cos u\sinh v$ follows from the exponential definition of sine. With $u=\pi /2$, its imaginary term vanishes, whileConsequently the real part is $5/4$ and the imaginary part is $0$.

Multiply numerator and denominator by the complex conjugate of the denominator:Squaring gives $(-1-3i{)}^2/4=(-8+6i)/4$. Thus the real part and imaginary part are

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.

Multiply by the complex conjugate of the denominator:Thus the real part and imaginary part are respectively

Multiply numerator and denominator by the complex conjugate of the denominator:Thus the real part and imaginary part are $\boxed{\mathrm{Re}=-5/13,\mathrm{Im}=-12/13}$.

The complex modulus is a nonnegative real number. Equality therefore requires $z-1$ itself to be real and nonnegative. Its imaginary part gives $y=0$, and its real part then gives $x-1\ge 0$. Conversely every such point satisfies the equality. Hence the complete solution set isGeometrically this is the closed ray on the real axis starting at $z=1$.

Use $|z{|}^2=z\bar{z}$ and $z-\bar{z}=2iy$. ThenThe real part and imaginary part are therefore

Expand the defining complex exponential functions:Thus the real part and imaginary part of the hyperbolic cosine are

A constant real part $\mathrm{Re}\ln z=L$ means $\ln |z|=L$, so its level sets are the concentric circlesTheir radii are strictly positive. The real part is independent of the logarithm's branch, so there is no branch ambiguity in these circles.

For $\mathrm{Re}(1/z)=K$, the equation is $x=K(x^2+y^2)$. If $K\ne 0$, complete the square to obtainThese circles have centres on the real axis and are tangent to the imaginary axis at the excluded origin. Positive $K$ puts the circle to the right, negative $K$ to the left. The separate case $K=0$ gives the punctured imaginary axis $\boxed{x=0,y\ne 0}$. The sketch uses an open marker at the origin to emphasize that $1/z$ is never defined there.

The complex exponential satisfies $e^{iz}=e^{-y}(\cos x+i\sin x)$. Multiplying by $x+iy$ givesThus its real part is

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

Put $c=\cos \theta$ and $s=\sin \theta$. Taking the real part of $(c+is{)}^6$ in De Moivre's theorem givesTaking the imaginary parts of the second and fourth powers gives $\sin 2\theta =2cs$ and $\sin 4\theta =4cs(c^2-s^2)$. HenceThe mixed powers cancel exactly.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.