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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 $z=1$.

Rewrite the rational function as $y=1-1/(x^3+1)$. It is undefined at $x=-1$, with the vertical asymptote $x=-1$, and tends to the horizontal asymptote $y=1$ as $x\to \pm \infty$. At the pole, $y\to +\infty$ from the left and $y\to -\infty$ from the right.

Its derivative isvanishing only at $x=0$. Since $y(0)=0$ and $y=x^3+O(x^6)$ near zero, this is a stationary inflection point, not an extremum. ThusThe graph increases on each side of the vertical asymptote. The left branch lies above $y=1$, while the right branch lies below it and crosses the origin horizontally.

The base $2e$ is constant. Write the exponential function as $\exp [\cos x\ln (2e)]$ and apply the chain rule:

Separate variables near the nonzero initial condition: $y^{-2}dy=dx$. Integration gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. ThereforeThis is the maximal connected interval containing the initial point. Direct differentiation gives $y^{\prime }=1/(2-x{)}^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical solution of a differential equation through that point.

In one period, sine equals $-\sqrt{3}/2$ at $4\pi /3$ and $5\pi /3$, and lies below that value between them. Since sine is a periodic function with period $2\pi$, all solutions are:The endpoints are included because the inequality is non-strict.