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Solve this equation:
[tex]\[
C_R = \{z \in \mathbb{C} \mid \Im(z) \ \textgreater \ 0 \ \text{and} \ 0 \ \textless \ |z| \ \textless \ R\}
\][/tex]


Sagot :

To solve the equation defined by [tex]\( C_R = \{ z \in \mathbb{C} \mid \Im(z) > 0 \text{ and } 0 < |z| < R \} \)[/tex], we need to identify the set of all complex numbers [tex]\( z \)[/tex] that satisfy the given conditions. Let's break down the conditions step-by-step.

### Step-by-Step Solution

1. Understanding the Complex Number [tex]\( z \)[/tex]:
A complex number [tex]\( z \in \mathbb{C} \)[/tex] can be represented in the form:
[tex]\[ z = x + yi \][/tex]
where [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are real numbers, and [tex]\( i \)[/tex] is the imaginary unit ([tex]\( i^2 = -1 \)[/tex]).

2. Condition 1: Imaginary Part Greater than 0:
The first condition [tex]\( \Im(z) > 0 \)[/tex] means that the imaginary part of [tex]\( z \)[/tex] must be positive. In terms of [tex]\( y \)[/tex], this translates to:
[tex]\[ y > 0 \][/tex]

3. Condition 2: Magnitude Less than R:
The second condition [tex]\( |z| < R \)[/tex] means that the modulus or magnitude of [tex]\( z \)[/tex] must be less than [tex]\( R \)[/tex]. The magnitude of [tex]\( z \)[/tex] is given by:
[tex]\[ |z| = \sqrt{x^2 + y^2} \][/tex]
Therefore, the condition [tex]\( |z| < R \)[/tex] translates to:
[tex]\[ \sqrt{x^2 + y^2} < R \][/tex]

4. Condition 3: Magnitude Greater than 0:
The third condition [tex]\( |z| > 0 \)[/tex] ensures that the magnitude of [tex]\( z \)[/tex] is positive. This eliminates the possibility of [tex]\( z \)[/tex] being the zero complex number. It is equivalent to:
[tex]\[ \sqrt{x^2 + y^2} > 0 \][/tex]

5. Combining the Conditions:
Combining all three conditions, we get:
[tex]\[ y > 0 \quad \text{(imaginary part positive)} \][/tex]
[tex]\[ \sqrt{x^2 + y^2} < R \quad \text{(magnitude less than R)} \][/tex]
[tex]\[ \sqrt{x^2 + y^2} > 0 \quad \text{(magnitude greater than 0)} \][/tex]

Since [tex]\( \sqrt{x^2 + y^2} > 0 \)[/tex] is always true for any non-zero complex number, this condition does not add any additional constraints beyond ensuring that [tex]\( z \neq 0 \)[/tex].

6. Visualizing the Solution in the Complex Plane:
In the complex plane, the set [tex]\( C_R \)[/tex] represents all points within a circle of radius [tex]\( R \)[/tex], centered at the origin, excluding the boundary (the circle itself) and the origin. Also, only the points in the upper half of the complex plane satisfy [tex]\( y > 0 \)[/tex].

Therefore, the complete set [tex]\( C_R \)[/tex] can be visualized as the interior of a semi-circle of radius [tex]\( R \)[/tex], centered at the origin, lying in the upper half of the complex plane (i.e., above the real axis), excluding the boundary.

7. Conclusion:
Hence, the set [tex]\( C_R \)[/tex] is:
[tex]\[ C_R = \{ z \in \mathbb{C} \mid \Im(z) > 0 \text{ and } 0 < |z| < R \} \][/tex]
This set consists of all complex numbers whose magnitude is less than [tex]\( R \)[/tex], excluding the origin, and whose imaginary part is positive.