<

Let's address each part of the question step-by-step, including graph sketches and analyses of the complex numbers, geometry, and verifying if the given curve is a hyperbola.

Part (a) $| z - i + 3 | = 1$

This equation is in the form $| z - c | = r$, which represents a circle in the complex plane. Here, $z = x + yi$, where $x$ is the real part and $y$ is the imaginary part of the complex number.

Solution steps:

1. The expression can be rewritten as $|z - (-3 + i)| = 1$.
2. This describes a circle with a center at $(-3, 1)$ in the complex plane (real part $-3$, imaginary part $1$).
3. The radius of the circle is $1$.

Summary for (a):

• The set of points is a circle centered at $(-3, 1)$ with a radius of $1$.
• This will be sketched as a circle in the complex plane, with several key points marked.

Part (b) $| 2z - 3i + 1 | \leq 2$

Here we analyze the inequality.

Solution steps:

1. Rewrite the equation in terms of $z = x + yi$: $|2(x + yi) - 3i + 1| \leq 2$ Simplify this: $|2x + (2y - 3)i + 1| \leq 2$
2. This inequality describes a disk centered at $(-1/2, 3/2)$ (real part $-\frac{1}{2}$, imaginary part $\frac{3}{2}$) with a radius of $1$.
3. The points satisfying this inequality form a closed disk in the complex plane.

Summary for (b):

• The set of points is a disk centered at $(-1/2, 3/2)$ with a radius of $1$, including the boundary of the disk.
• This will be sketched as a filled circle.

Part (c) $z^2 + \overline{z^2} = 2$

This equation involves both $z^2$ and its complex conjugate $\overline{z^2}$.

Solution steps:

1. Let $z = x + yi$, where $x$ and $y$ are real and imaginary parts of $z$, respectively.
2. Compute $z^2 = (x + yi)^2 = x^2 - y^2 + 2xyi$, and $\overline{z^2} = (x^2 - y^2 - 2xyi)$.
3. Add $z^2$ and $\overline{z^2}$: $z^2 + \overline{z^2} = (x^2 - y^2 + 2xyi) + (x^2 - y^2 - 2xyi) = 2(x^2 - y^2)$
4. Set this equal to 2: $2(x^2 - y^2) = 2$ Dividing both sides by 2: $x^2 - y^2 = 1$
5. This is the standard equation of a hyperbola.

Summary for (c):

• The equation $z^2 + \overline{z^2} = 2$ represents a hyperbola with the equation $x^2 - y^2 = 1$.
• The graph will be a hyperbola in the complex plane.

---

Verifying Hyperbola:

We derived $x^2 - y^2 = 1$, which is the equation of a hyperbola, confirming that the equation represents a hyperbola in the complex plane.

Sketching the Sets:

1. For part (a), a circle centered at $(-3, 1)$ with radius 1.
2. For part (b), a disk centered at $(-1/2, 3/2)$ with radius 1.
3. For part (c), a hyperbola given by $x^2 - y^2 = 1$, symmetric about the x-axis.

Would you like to see the sketch of the graphs?

---

5 Related Questions:

1. How do you convert a complex number into polar form?
2. How do you derive the equation of a circle in the complex plane?
3. What is the significance of the magnitude of a complex number in geometry?
4. Can complex conjugates be used to simplify equations involving imaginary parts?
5. What are the geometric interpretations of inequalities involving complex numbers?

Tip:

To verify if a given curve is a hyperbola or another conic section, look for terms like $x^2 - y^2$ in the equation.