z satisfies mod(z-2-2i)=2 and arg(z-2-2i)=pi/6, find z

We are tasked with finding a complex number \( z \) that satisfies the following two conditions:

1. \( |z - (2 + 2i)| = 2 \)
2. \( \arg(z - (2 + 2i)) = \frac{\pi}{6} \)

### Step 1: Representing \( z \) in terms of a shifted complex number

Let:
\[
 w = z - (2 + 2i)
\]
This means we are looking for \( w \) such that:

- \( |w| = 2 \)
- \( \arg(w) = \frac{\pi}{6} \)

### Step 2: Using the polar form of complex numbers

From polar coordinates, we know that any complex number \( w \) can be expressed as:
\[
 w = r(\cos \theta + i \sin \theta)
\]
where \( r \) is the magnitude and \( \theta \) is the argument of \( w \). 

Given that \( |w| = 2 \) and \( \arg(w) = \frac{\pi}{6} \), we have:
\[
 w = 2\left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right)
\]

### Step 3: Simplifying the expression for \( w \)

We now calculate the trigonometric values:
\[
 \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin\frac{\pi}{6} = \frac{1}{2}
\]
Thus, \( w \) becomes:
\[
 w = 2\left( \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right)
\]
\[
 w = \sqrt{3} + i
\]

### Step 4: Finding \( z \)

Since \( w = z - (2 + 2i) \), we have:
\[
 z - (2 + 2i) = \sqrt{3} + i
\]
Solving for \( z \):
\[
 z = (\sqrt{3} + i) + (2 + 2i)
\]
\[
 z = \sqrt{3} + 2 + (1 + 2)i
\]
\[
 z = (\sqrt{3} + 2) + (3i)
\]

Thus, the complex number \( z \) is:
\[
 z = \sqrt{3} + 2 + 3i
\]

### Final Answer:
\[
 z = \sqrt{3} + 2 + 3i
\]

Would you like further details or have any questions?

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Here are 5 related questions:
1. How do we find the modulus of a complex number?
2. What is the geometric interpretation of the argument of a complex number?
3. How can we express a complex number in polar form?
4. How does adding a complex number shift it in the complex plane?
5. What is the connection between Euler's formula and polar representation?

**Tip:** When solving complex number problems, converting between rectangular and polar forms can simplify operations like multiplication, division, and finding powers or roots.

The query submitted by the user: z satisfies mod(z-2-2i)=2 and arg(z-2-2i)=pi/6, find z

Solve the complex number problem where z satisfies |z - (2 + 2i)| = 2 and arg(z - (2 + 2i)) = π/6. Learn how to use polar coordinates, modulus, and argument to find z. This solution is perfect for students in Grades 10-12 and covers core concepts in complex numbers.

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