Where do the points lie in the complex plane for which the…
Where do the points lie in the complex plane for which the following conditions are satisfied: $$ \begin{aligned} & 1^{\circ}\ \ 2<|z|<3,\ \ \frac{\pi}{4}<\varphi<\frac{\pi}{3}; \quad 2^{\circ}\ \ |z|>1,\ \ \frac{\pi}{3}<\varphi<\frac{\pi}{2}; \\ & 3^{\circ}\ \ |z-z_0|<1; \quad 4^{\circ}\ \ |z-i|<1\ \text{and}\ |z-1|<1; \quad 5^{\circ}\ \ |z|+\text{Re}(z)<2. \end{aligned} $$
Explanation of Method
Inequalities on $|z|$ describe regions bounded by circles (or annuli), inequalities on $\varphi = \arg z$ describe angular sectors, and inequalities on $|z-z_0|$ describe disks centered at $z_0$. Combined conditions ("and") describe the intersection of the individual regions. When an inequality mixes $|z|$ and $\text{Re}(z)$, substitute $z = x+iy$, isolate the square root, and square carefully (respecting the sign needed for squaring to be valid) to reveal the boundary curve.
Part 1° · $2 < |z| < 3,\ \ \dfrac{\pi}{4} < \varphi < \dfrac{\pi}{3}$
The modulus condition $2 < |z| < 3$ restricts points to the region strictly between the circles $|z|=2$ and $|z|=3$. The argument condition $\dfrac{\pi}{4} < \varphi < \dfrac{\pi}{3}$ (45° to 60°) restricts points to the angular sector between those two rays, in Quadrant I.
$$ \boxed{\text{annular sector: } 2<|z|<3 \text{ with } 45° < \varphi < 60°} $$
The boundary consists of the inner circle $|z|=2$, the outer circle $|z|=3$, and the two rays at angles $\frac{\pi}{4}$ and $\frac{\pi}{3}$ (none included, since all inequalities are strict).
Annular sector: between circles $|z|=2$ and $|z|=3$, angles $45°$ to $60°$.
Part 2° · $|z| > 1,\ \ \dfrac{\pi}{3} < \varphi < \dfrac{\pi}{2}$
The condition $|z|>1$ excludes the closed unit disk, leaving everything outside the unit circle. Combined with $\dfrac{\pi}{3} < \varphi < \dfrac{\pi}{2}$ (60° to 90°), the region is an unbounded angular sector.
$$ \boxed{\text{unbounded sector: } |z|>1 \text{ with } 60° < \varphi < 90°} $$
It is bounded by the unit circle $|z|=1$, the ray at angle $\frac{\pi}{3}$, and the ray at angle $\frac{\pi}{2}$ (the positive imaginary axis), extending outward to infinity.
Unbounded sector outside $|z|=1$, angles $60°$ to $90°$ (shaded portion shown; region continues outward).
Part 3° · $|z - z_0| < 1$
The expression $|z-z_0|$ is the distance from $z$ to the fixed point $z_0$. The condition $|z-z_0|<1$ says that distance is less than 1.
$$ \boxed{\text{open disk of radius } 1 \text{ centered at } z_0} $$
If $z_0 = x_0 + iy_0$, the region is $\{(x,y) : (x-x_0)^2 + (y-y_0)^2 < 1\}$ — the interior of the circle, not including the boundary.
Open disk of radius $1$ centered at an arbitrary point $z_0$ (dashed boundary = excluded).
Part 4° · $|z - i| < 1$ and $|z - 1| < 1$
This is the simultaneous (intersection) of two open unit disks: one centered at $i = (0,1)$, the other centered at $1 = (1,0)$.
For $z = x+iy$: $|z-i|^2<1 \Rightarrow x^2+(y-1)^2<1$, and $|z-1|^2<1 \Rightarrow (x-1)^2+y^2<1$.
The distance between the two centers is $|i-1| = \sqrt{1^2+1^2} = \sqrt{2} \approx 1.414$. Since $\sqrt{2} < 1+1 = 2$ (sum of the radii), the circles overlap, and the intersection is a nonempty lens. Solving the two circle equations simultaneously shows the circles meet exactly at $z=0$ and $z=1+i$.
$$ \boxed{\text{lens-shaped intersection of the two open unit disks centered at } i \text{ and } 1} $$
Lens intersection of the unit disks centered at $i$ and $1$, meeting at $z=0$ and $z=1+i$.
Part 5° · $|z| + \text{Re}(z) < 2$
Let $z = x+iy$. The condition becomes $\sqrt{x^2+y^2} + x < 2$, i.e. $\sqrt{x^2+y^2} < 2-x$.
For this to be meaningful we need $2-x>0$, i.e. $x<2$. Squaring both sides (valid since $2-x>0$):
$$ x^2+y^2 < (2-x)^2 = 4-4x+x^2 \quad\Longrightarrow\quad y^2 < 4-4x \quad\Longrightarrow\quad y^2 < 4(1-x) $$
Rewriting as $y^2 = -4(x-1)$ shows the boundary is a parabola opening to the left with vertex at $(1,0)$, and the region is its interior (the concave side, toward $x \to -\infty$).
$$ \boxed{y^2 < 4(1-x), \quad \text{i.e. the interior of the parabola } y^2 = 4(1-x) \text{ with vertex } (1,0)} $$
Interior of the parabola $y^2 = 4(1-x)$, vertex at $(1,0)$, opening to the left.