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Set $w=e^z\ne 0$. The hyperbolic cosine equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two polynomial roots are $w=-2\pm \sqrt{3}$, both negative. Taking every value of the complex logarithm yieldsSince $(2-\sqrt{3})(2+\sqrt{3})=1$, their real logarithms have opposite signs. All solutions are thereforeThe positive real constant is also $\mathrm{arcosh}2$, the inverse hyperbolic cosine of $2$. Substitution gives $\cosh z=-\cosh (\ln (2+\sqrt{3}))=-2$ for both signs and every integer $k$.

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.

In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, with $r=\sqrt{x^2+y^2}$. Thus the polar form of a complex number isHere $r=|z|\ge 0$ is its complex modulus; for $z\ne 0$, the complex argument $\theta$ is determined modulo $2\pi$. At $z=0$ the angle is undefined, although every value of $\theta$ gives the same point when $r=0$.

The dot product formula for the angle between two nonzero vectors gives

The Euclidean norm of each position vector gives

The separable differential equation gives $dy/y=-dx$, so $y=C{e}^{-x}$. The initial condition fixes $C=2$, and hence

On the negative half-line, an antiderivative of $1/x$ is $\ln |x|$. The fundamental theorem of calculus therefore givesThe negative sign agrees with the negative integrand throughout the interval.

Differentiation of the finite sum involves no convergence issue. The derivative has the Fourier cosine series

Use the double-angle identity $2\sin x\cos x=\sin 2x$. The Fourier series is already a finite trigonometric sum:In the usual Fourier sine series convention, $b_1=b_2=b_3=1$, all other $b_n$ vanish, and all cosine coefficients and the constant coefficient vanish.

The fundamental theorem for line integrals evaluates this conservative vector field integral by the change in potential. At the initial point $\varphi =0$. At the final point $z=0$, so the exponential factor is one, andThe detailed shape of the path does not affect this line integral.

The potential is smooth everywhere, so its mixed partial derivatives commute. The curl of a gradient therefore vanishes: