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Euler's formula gives $e^{iz}=e^{-y}(\cos x+i\sin x)$. Its real part and imaginary part are

Multiply numerator and denominator by the complex conjugate of $1+z$. For $z\ne -1$ this givesThus its real part and imaginary part are respectively

Writing $z=x+iy$ and comparing real parts and imaginary parts gives $\sin x\cosh y=\cosh 2$ and $\cos x\sinh y=0$. A real solution $y=0$ is impossible since $\cosh 2>1$. Thus $\cos x=0$, and positivity forces $\sin x=1$. Consequently $x=\pi /2+2k\pi$ and $\cosh y=\cosh 2$, givingThe right Argand diagram in the preceding figure shows representative points on the lines $\mathrm{Im}z=\pm 2$; each row continues with spacing $2\pi$.

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

Taking the coefficient of $i$ in the expansion from part (a), the imaginary part is

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.