Okay, so I need to find z_2002 where each term is generated by applying the function F(z) to the previous term. The function F(z) is given as (z + i)/(z - i), and the starting value z_0 is 1/137 + i. Alright, let's start by understanding the problem step by step.

First, maybe I should compute the first few terms to see if there's a pattern. If there's a cycle or periodicity, then I can find the period and use that to compute z_2002 without calculating all terms up to 2002. That would save a lot of time.

Let me compute z_1 first. Using the formula z_n = F(z_{n-1}), so z_1 = F(z_0). Let's substitute z_0 into F(z):

z_1 = F(z_0) = [ (1/137 + i) + i ] / [ (1/137 + i) - i ]

Simplify numerator and denominator:

Numerator: (1/137 + i) + i = 1/137 + 2i

Denominator: (1/137 + i) - i = 1/137

So z_1 = (1/137 + 2i) / (1/137) = [1/137]/[1/137] + [2i]/[1/137] = 1 + 2i * 137 = 1 + 274i

So z_1 is 1 + 274i. Hmm, okay. Let's compute z_2 next:

z_2 = F(z_1) = (1 + 274i + i) / (1 + 274i - i) = (1 + 275i) / (1 + 273i)

Now, to simplify this complex fraction, multiply numerator and denominator by the conjugate of the denominator:

Numerator: (1 + 275i)(1 - 273i) = 1*(1) + 1*(-273i) + 275i*1 + 275i*(-273i)

= 1 - 273i + 275i - 275*273 i^2

Since i^2 = -1, so the last term becomes +275*273

Combine like terms:

Real parts: 1 + 275*273

Imaginary parts: (-273i + 275i) = 2i

So real part: 1 + 275*273. Let me compute 275*273. 275*200 = 55,000; 275*73 = 275*(70 + 3) = 275*70 = 19,250; 275*3 = 825. So 19,250 + 825 = 20,075. Then total real part is 1 + 20,075 = 20,076.

Imaginary part: 2i

Denominator: (1 + 273i)(1 - 273i) = 1^2 - (273i)^2 = 1 - (273)^2 * i^2 = 1 - (273)^2*(-1) = 1 + (273)^2

Compute 273^2: 273*273. Let's compute 270^2 = 72,900; 2*270*3 = 1,620; 3^2=9. So (270 + 3)^2 = 72,900 + 1,620 + 9 = 74,529.

So denominator is 1 + 74,529 = 74,530.

Therefore, z_2 = (20,076 + 2i)/74,530. Let's simplify this by dividing numerator and denominator by 2:

Numerator: 10,038 + i

Denominator: 37,265

So z_2 = (10,038 + i)/37,265. Hmm, this is getting messy. Maybe there's a pattern here, but it's not obvious yet. Let's compute z_3 to see.

z_3 = F(z_2) = (z_2 + i)/(z_2 - i)

Substituting z_2 = (10,038 + i)/37,265:

Numerator: [(10,038 + i)/37,265] + i = (10,038 + i + 37,265i)/37,265 = (10,038 + 37,266i)/37,265

Denominator: [(10,038 + i)/37,265] - i = (10,038 + i - 37,265i)/37,265 = (10,038 - 37,264i)/37,265

So z_3 is [ (10,038 + 37,266i)/37,265 ] / [ (10,038 - 37,264i)/37,265 ] = (10,038 + 37,266i)/(10,038 - 37,264i)

Multiply numerator and denominator by the conjugate of the denominator:

Numerator: (10,038 + 37,266i)(10,038 + 37,264i)

Denominator: (10,038)^2 + (37,264)^2

This seems quite complex. Maybe there's a different approach here. Instead of computing each term step by step, perhaps there's a functional equation or a cyclic behavior we can exploit.

Let me consider the function F(z). Maybe it's a Möbius transformation, and such transformations have properties that can lead to periodic cycles. If F is a rotation or inversion, applying it multiple times might cycle through a finite set of points.

Alternatively, perhaps we can represent F(z) in terms of a matrix and see if the transformation has finite order. Möbius transformations can be represented by matrices, and if the matrix has finite order, then applying the transformation repeatedly cycles through a finite number of transformations.

Let me recall that a Möbius transformation is given by f(z) = (az + b)/(cz + d), and the corresponding matrix is [[a, b], [c, d]]. The composition of transformations corresponds to matrix multiplication. If the matrix has finite order, then after some number of compositions, the transformation is the identity, leading to periodicity.

So, for our function F(z) = (z + i)/(z - i), the corresponding matrix is:

[[1, i],
 [1, -i]]

Let me check that. The general form is (az + b)/(cz + d), so a=1, b=i, c=1, d=-i. So matrix is:

[1   i]
[1  -i]

Now, to see if this matrix has finite order, let's compute its powers. If there exists an integer k such that the matrix raised to the k-th power is a scalar multiple of the identity matrix, then the transformation would have finite order.

Alternatively, we can compute the trace and determinant to see the eigenvalues. The trace is 1 + (-i) = 1 - i. The determinant is (1)(-i) - (i)(1) = -i - i = -2i.

Eigenvalues satisfy λ^2 - (trace)λ + determinant = 0. So:

λ^2 - (1 - i)λ - 2i = 0

Compute the discriminant: (1 - i)^2 + 8i = (1 - 2i + i^2) + 8i = (1 - 2i -1) + 8i = (-2i) + 8i = 6i

So eigenvalues are [ (1 - i) ± sqrt(6i) ] / 2

Hmm, sqrt(6i) is a bit complicated. Let me compute sqrt(6i). Let’s write 6i in polar form. 6i has magnitude 6 and angle π/2. So sqrt(6i) is sqrt(6) e^{iπ/4} = sqrt(6)(cos π/4 + i sin π/4) = sqrt(6)*(√2/2 + i√2/2) = sqrt(3) + i sqrt(3)

Therefore, sqrt(6i) = sqrt(3) + i sqrt(3). So the eigenvalues are:

[ (1 - i) ± (sqrt(3) + i sqrt(3)) ] / 2

Let’s compute both possibilities:

First eigenvalue: [1 - i + sqrt(3) + i sqrt(3)] / 2 = [1 + sqrt(3) + i(-1 + sqrt(3))]/2

Second eigenvalue: [1 - i - sqrt(3) - i sqrt(3)] / 2 = [1 - sqrt(3) + i(-1 - sqrt(3))]/2

These eigenvalues don't look like roots of unity, which would be necessary for the matrix to have finite order. Therefore, perhaps the transformation doesn't have finite order. Then maybe the sequence z_n cycles with some period, but not due to the transformation's order. Alternatively, maybe there's another approach.

Alternatively, perhaps we can find a fixed point of F(z) and conjugate the function to a rotation. Fixed points are solutions to F(z) = z.

Set (z + i)/(z - i) = z

Multiply both sides by (z - i):

z + i = z(z - i)

Expand right side: z^2 - iz

Bring all terms to left:

z + i - z^2 + iz = 0

Rearranged: -z^2 + (1 + i)z + i = 0

Multiply by -1: z^2 - (1 + i)z - i = 0

Use quadratic formula:

z = [ (1 + i) ± sqrt( (1 + i)^2 + 4i ) ] / 2

Compute discriminant:

(1 + i)^2 + 4i = 1 + 2i + i^2 + 4i = 1 + 2i -1 + 4i = 6i

So sqrt(6i) is sqrt(3) + i sqrt(3), as before.

Therefore,

z = [ (1 + i) ± (sqrt(3) + i sqrt(3)) ] / 2

So the fixed points are:

First solution: [1 + i + sqrt(3) + i sqrt(3)] / 2 = [1 + sqrt(3) + i(1 + sqrt(3))]/2

Second solution: [1 + i - sqrt(3) - i sqrt(3)] / 2 = [1 - sqrt(3) + i(1 - sqrt(3))]/2

So the fixed points are (1 + sqrt(3))/2 + i(1 + sqrt(3))/2 and (1 - sqrt(3))/2 + i(1 - sqrt(3))/2.

These are complex numbers. Let's denote them by α and β.

Now, if we conjugate the function F(z) using a Möbius transformation that maps α to 0 and β to infinity, then the transformed function would be a rotation. Then, applying F multiple times would correspond to rotating around the origin, which has finite order if the rotation angle is a rational multiple of π.

But this seems a bit involved. Let me see if there's another way.

Alternatively, maybe we can consider the transformation F(z) as a rotation in the complex plane. If we can express F(z) as a rotation composed with inversion or something, then applying it multiple times could result in a rotation by a fixed angle each time. Then, after 2002 applications, we can compute the total rotation.

Let me see. Let's write F(z) in terms of complex numbers. Let’s note that F(z) = (z + i)/(z - i). Let me see if this can be expressed as a rotation.

Alternatively, perhaps we can write F(z) as a transformation that maps the upper half-plane to the unit disk or something. Wait, maybe not. Let's try another approach.

Suppose we consider the transformation F(z) = (z + i)/(z - i). Let's compute F(z) when z is on the imaginary axis. Let z = iy. Then F(iy) = (iy + i)/(iy - i) = i(y + 1)/i(y - 1) = (y + 1)/(y - 1). So it maps the imaginary axis to the real axis. Interesting. Similarly, if we take z = x + iy, perhaps we can express F(z) in terms of real and imaginary parts.

Alternatively, let's compute the effect of F(z) on the complex plane. The function F(z) is a Möbius transformation that maps i to infinity and -i to 0. Wait, let's check:

F(i) = (i + i)/(i - i) = (2i)/0, which is infinity. So F(z) maps i to infinity. Similarly, F(-i) = (-i + i)/(-i - i) = 0/(-2i) = 0. So F(z) maps -i to 0. Therefore, this Möbius transformation has two fixed points: the points that are invariant under F(z). Wait, but earlier we found fixed points. Let me check if the fixed points are correct.

Wait, earlier when solving F(z) = z, we found two fixed points, which are α and β. Therefore, the Möbius transformation F(z) has two fixed points. Möbius transformations with two fixed points are either parabolic, hyperbolic, or elliptic, depending on their trace. Since the trace squared is (1 - i)^2 = 1 - 2i -1 = -2i, which isn't a real number. Hmm, maybe this is getting too complex.

Alternatively, perhaps we can consider the transformation F(z) in terms of its action on the Riemann sphere. Möbius transformations can be classified based on their fixed points and their multipliers (the eigenvalues of the corresponding matrix). Since the eigenvalues we found earlier were complex and not roots of unity, maybe the transformation is of hyperbolic type, leading to divergent behavior. However, since we are dealing with complex dynamics, perhaps periodic points emerge. But perhaps there's a different angle.

Wait, maybe we can express z_n in terms of z_0 using the function F composed n times. So z_n = F^n(z_0), where F^n is the n-th iterate of F. If we can find a formula for F^n(z), then we can plug in z_0 and compute z_2002.

Alternatively, perhaps we can observe that F is an involution, meaning that F(F(z)) = z. Let me check that:

Compute F(F(z)):

F(F(z)) = F( (z + i)/(z - i) ) = [ ( (z + i)/(z - i) ) + i ] / [ ( (z + i)/(z - i) ) - i ]

Simplify numerator and denominator:

Numerator: [ (z + i) + i(z - i) ] / (z - i)

Denominator: [ (z + i) - i(z - i) ] / (z - i)

So overall, F(F(z)) = [ (z + i + iz - i^2) / (z - i) ] / [ (z + i - iz + i^2) / (z - i) ] = [ (z + i + iz + 1) ] / [ (z + i - iz - 1) ]

Simplify numerator and denominator:

Numerator: z + i + iz + 1 = (1 + i)z + (1 + i) = (1 + i)(z + 1)

Denominator: z + i - iz -1 = (1 - i)z + (i -1) = (1 - i)z + (i -1) = (1 - i)(z -1)

So F(F(z)) = [ (1 + i)(z +1) ] / [ (1 - i)(z -1) ]

Multiply numerator and denominator by (1 + i)/(1 + i):

Numerator: (1 + i)^2 (z +1) = (1 + 2i + i^2)(z +1) = (1 + 2i -1)(z +1) = 2i(z +1)

Denominator: (1 - i)(1 + i)(z -1) = (1 - i^2)(z -1) = (1 +1)(z -1) = 2(z -1)

Therefore, F(F(z)) = [2i(z +1)] / [2(z -1)] = i(z +1)/(z -1)

Hmm, not equal to z. So F(F(z)) is not equal to z. Therefore, F is not an involution.

Wait, but maybe F^4(z) = z? Let's check.

Compute F^3(z) = F(F(F(z))). Since F^2(z) = i(z +1)/(z -1), then F^3(z) = F(F^2(z)) = F( i(z +1)/(z -1) )

Compute that:

F(w) where w = i(z +1)/(z -1)

So F(w) = (w + i)/(w - i) = [ i(z +1)/(z -1) + i ] / [ i(z +1)/(z -1) - i ]

Simplify numerator and denominator:

Numerator: [ i(z +1) + i(z -1) ] / (z -1 ) = [ i(z +1 + z -1) ] / (z -1 ) = [ i(2z) ] / (z -1 ) = 2i z / (z -1 )

Denominator: [ i(z +1) - i(z -1) ] / (z -1 ) = [ i(z +1 - z +1) ] / (z -1 ) = [ i(2) ] / (z -1 ) = 2i / (z -1 )

Therefore, F^3(z) = [2i z / (z -1 ) ] / [2i / (z -1 ) ] = z

So F^3(z) = z. Therefore, applying F three times brings us back to the original z. Therefore, F is periodic with period 3. So F^3(z) = z for all z ≠ i. Therefore, the transformation F has order 3. Therefore, applying F three times returns to the original value. Therefore, the sequence z_n is periodic with period 3. Therefore, z_{n} = z_{n mod 3}.

Therefore, to find z_{2002}, compute 2002 mod 3.

Compute 2002 divided by 3: 3*667 = 2001, so 2002 = 3*667 +1, so 2002 mod3 =1. Therefore, z_{2002}= z_1.

Earlier, we computed z_1 = 1 + 274i. Therefore, z_{2002}=1 +274i.

Wait, is that correct? Let me verify.

Wait, from the computation, F^3(z) = z. Therefore, the sequence cycles every 3 steps. Therefore, z_0, z_1, z_2, z_3 = z_0, so cycle length is 3. Therefore, z_n = z_{n mod3}.

Therefore, z_{2002}= z_{2002 mod3}. Compute 2002 divided by3: 3*667=2001, so 2002=2001+1, so 2002 mod3=1. Therefore, z_{2002}= z_1=1 +274i.

But in the earlier computation, z_1 was 1 +274i, which is correct. Then z_2 was (10,038 +i)/37,265, which is approximately 0.27 +0.000027i. Then z_3 would be z_0 again. So the cycle is z_0 -> z_1 -> z_2 -> z_0. Therefore, the period is 3. Therefore, yes, z_{2002}=z_1.

Therefore, the answer is 1 +274i. However, let me check once again.

Wait, if F^3(z)=z, then z_3 = z_0. So z_3 = z_0, z_4 = z_1, z_5 = z_2, z_6 = z_3 = z_0, etc. Therefore, z_{3k}=z_0, z_{3k+1}=z_1, z_{3k+2}=z_2.

Therefore, 2002 divided by3: 2002=3*667 +1, so 2002=3*667 +1, so 2002=3k+1, where k=667. Therefore, z_{2002}=z_1=1 +274i.

Therefore, the answer is 1 +274i. But let me check the calculation of z_1 again to be sure.

z_1 = F(z_0) = (z_0 +i)/(z_0 -i) with z_0=1/137 +i

So numerator: (1/137 +i) +i =1/137 +2i

Denominator: (1/137 +i) -i =1/137

Therefore, z_1 = (1/137 +2i)/(1/137) =1 +274i. Correct.

Therefore, the answer is z_1=1 +274i. So the final answer is \boxed{1 + 274i}

**Final Answer**
\boxed{1 + 274i}
Therefore, the value of \( z_{2002} \) is \( \boxed{1 + 274i} \).
</think>