Okay, so I need to find the complex number w, which is the result when z is rotated around c by π/4 counter-clockwise. Let me start by recalling how rotations work in the complex plane. 

First, I remember that rotating a point around another point involves translating the system so that the center of rotation becomes the origin, applying the rotation, and then translating back. The formula for this is something like: if you want to rotate a complex number z around a point c by an angle θ, you translate z by subtracting c, rotate it by multiplying by e^{iθ}, and then add c back. 

So, mathematically, that would be w = (z - c) * e^{iθ} + c. Let me check if that makes sense. If we subtract c from z, we move the center of rotation to the origin. Then multiplying by e^{iθ} rotates the vector from c to z by θ. Then adding c back translates it back to the original position. Yeah, that seems right.

Given that, let's write down the values. z is given as 2 + √2 - (3 + 3√2)i. So in complex form, z = (2 + √2) + (-3 - 3√2)i. And c is 2 - 3i. So c = 2 - 3i. 

First, compute z - c. Let's subtract the real parts and the imaginary parts separately. 

Real part: (2 + √2) - 2 = √2. 

Imaginary part: (-3 - 3√2) - (-3) = -3 - 3√2 + 3 = -3√2. 

So z - c = √2 - 3√2 i. 

Wait, that seems straightforward. Then, we need to multiply this by e^{iπ/4}, which is the rotation factor for 45 degrees counter-clockwise. 

e^{iπ/4} is cos(π/4) + i sin(π/4) = √2/2 + i√2/2. 

Therefore, multiplying (z - c) by e^{iπ/4} would be:

(√2 - 3√2 i) * (√2/2 + i√2/2). 

Let me compute this multiplication step by step. Let's denote the first complex number as a + bi, where a = √2 and b = -3√2. The second complex number is c + di, where c = √2/2 and d = √2/2. 

Multiplying them: (a + bi)(c + di) = (ac - bd) + (ad + bc)i. 

So calculating the real part: ac - bd = √2*(√2/2) - (-3√2)*(√2/2). 

First term: √2 * √2/2 = 2/2 = 1. 

Second term: -(-3√2)*(√2/2) = 3√2 * √2/2 = 3*(2)/2 = 3. 

So real part: 1 + 3 = 4. 

Imaginary part: ad + bc = √2*(√2/2) + (-3√2)*(√2/2). 

First term: √2 * √2/2 = 2/2 = 1. 

Second term: -3√2 * √2/2 = -3*(2)/2 = -3. 

So imaginary part: 1 - 3 = -2. 

Therefore, the product (z - c) * e^{iπ/4} is 4 - 2i. 

Then, to get w, we add c back. So w = (4 - 2i) + (2 - 3i) = (4 + 2) + (-2 - 3)i = 6 - 5i. 

Wait, that seems straightforward, but let me verify each step again to make sure I didn't make any calculation errors.

Starting with z - c: z = 2 + √2 - (3 + 3√2)i, c = 2 - 3i. 

Subtracting c from z: (2 + √2 - 2) + (-3 - 3√2 + 3)i = √2 - 3√2 i. Correct.

Multiplying by e^{iπ/4}: √2 - 3√2 i multiplied by √2/2 + √2/2 i. Let's do this multiplication again step by step.

First, expand the product:

(√2)(√2/2) + (√2)(√2/2 i) + (-3√2 i)(√2/2) + (-3√2 i)(√2/2 i)

Compute each term:

First term: (√2)(√2/2) = (2)/2 = 1.

Second term: (√2)(√2/2 i) = (2)/2 i = i.

Third term: (-3√2 i)(√2/2) = (-3√2 * √2)/2 i = (-3*2)/2 i = -3i.

Fourth term: (-3√2 i)(√2/2 i) = (-3*2)/2 * i^2 = (-3) * (-1) = 3.

Combine all terms:

1 + i - 3i + 3 = (1 + 3) + (i - 3i) = 4 - 2i. 

Yes, that's correct. Then adding c: 4 - 2i + 2 - 3i = 6 - 5i. 

So w = 6 - 5i. 

But wait, let me check once again. Maybe I made a mistake in the multiplication. Let me recompute the multiplication using the distributive property.

Let me denote (a + bi)(c + di) = ac + adi + bci + bdi^2. Since i^2 = -1, this becomes (ac - bd) + (ad + bc)i. 

Applying this formula:

a = √2, b = -3√2, c = √2/2, d = √2/2.

Real part: ac - bd = √2*(√2/2) - (-3√2)*(√2/2) = (2/2) - (-3*2/2) = 1 - (-3) = 1 + 3 = 4. Correct.

Imaginary part: ad + bc = √2*(√2/2) + (-3√2)*(√2/2) = (2/2) + (-3*2/2) = 1 - 3 = -2. Correct.

So, the product is indeed 4 - 2i. Adding c: 4 - 2i + 2 - 3i = 6 - 5i. 

Therefore, the final answer should be 6 - 5i. 

But let me think again: is there another way to represent the rotation? For example, using complex numbers in polar form or something else?

Alternatively, since rotation around a point can be represented as:

w = c + (z - c) * e^{iθ}

Which is the same as what I did before. So, substituting θ = π/4, which is 45 degrees. 

Alternatively, perhaps using the formula for rotation in terms of vectors. But since we're dealing with complex numbers, the method I used is appropriate. 

Alternatively, maybe represent z and c in coordinate form and compute the rotation via vectors. Let's check that.

z is (2 + √2, -3 - 3√2). c is (2, -3). So the vector from c to z is (2 + √2 - 2, -3 - 3√2 - (-3)) = (√2, -3√2). 

So the vector is (√2, -3√2). Rotating this vector by π/4 counter-clockwise. The rotation matrix for π/4 is:

[cos(π/4) -sin(π/4)]
[sin(π/4) cos(π/4)]

Which is:

[√2/2  -√2/2]
[√2/2   √2/2]

Multiplying this matrix by the vector (√2, -3√2):

First component: √2/2 * √2 + (-√2/2) * (-3√2) 

= (2)/2 + (3*2)/2 

= 1 + 3 = 4.

Second component: √2/2 * √2 + √2/2 * (-3√2)

= (2)/2 + (-3*2)/2 

= 1 - 3 = -2.

So the rotated vector is (4, -2). Then, adding back the point c (2, -3), the new point w is (4 + 2, -2 + (-3)) = (6, -5). 

Which corresponds to the complex number 6 - 5i. 

Therefore, this method also gives the same result. So that confirms that the answer is indeed 6 - 5i. 

Therefore, after verifying through two different methods (direct complex multiplication and vector rotation with rotation matrix), I'm confident that the result is correct. 

I don't see any mistakes in the calculations. Both approaches lead to the same conclusion, so I think that's solid. 

**Final Answer**
The result when $z$ is rotated around $c$ by $\frac{\pi}{4}$ counter-clockwise is $\boxed{6 - 5i}$.
Therefore, the result when $z$ is rotated around $c$ by $\frac{\pi}{4}$ counter-clockwise is $\boxed{6 - 5i}$.
</think>