<Question>
Let $z$ be a complex number such that $|z| = 1.$  Find the maximum value of
\[|1 + z| + |1 - z + z^2|.\]
</Question>

<think>
<Plan_1>
Okay, so I need to find the maximum value of the expression |1 + z| + |1 - z + z²| where z is a complex number with |z| = 1. Hmm, let's start by understanding what's given here.

<prompt_1.1> Understand the given conditions and express z in terms of trigonometric functions. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, since |z| = 1, z lies on the unit circle in the complex plane. That means I can represent z as e^{iθ} where θ is the argument of z. Alternatively, z can be written as cosθ + i sinθ. Maybe using the exponential form will make things easier when dealing with powers of z.

Let me write z = e^{iθ}. Then z² = e^{i2θ}. That should help in computing the modulus expressions.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let's break down the expression into two parts: |1 + z| and |1 - z + z²|. I need to compute each modulus separately and then add them together, then find the maximum value of the sum.

Starting with |1 + z|.

Since z = e^{iθ}, 1 + z = 1 + e^{iθ}.

<prompt_2.1> Compute the modulus |1 + z| using trigonometric identities. </prompt_2.1>

Now the second part: |1 - z + z²|.

Again, substituting z = e^{iθ}, so z² = e^{i2θ}.

<prompt_2.2> Compute the modulus |1 - z + z²| using trigonometric identities. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
I can use the formula for the modulus of a complex number: |a + ib| = √(a² + b²). Alternatively, there's a trigonometric identity for |1 + e^{iθ}|. Let me recall that. 

Yes, |1 + e^{iθ}| = 2|cos(θ/2)|. Let me verify that. 

Expressed in terms of cosine, the modulus squared would be (1 + cosθ)² + (sinθ)² = 1 + 2cosθ + cos²θ + sin²θ = 2 + 2cosθ. Therefore, modulus is √(2 + 2cosθ) = √(4cos²(θ/2)) = 2|cos(θ/2)|. Since θ is between 0 and 2π, cos(θ/2) can be positive or negative, but the modulus is always non-negative. So |1 + z| = 2|cos(θ/2)|.
</execution_2.1>

<execution_2.2>
So 1 - z + z² = 1 - e^{iθ} + e^{i2θ}.

Hmm, perhaps I can also express this in terms of sine and cosine. Let me try expanding both moduli.

First, |1 + z|.

Expressed in terms of cosine:

1 + z = 1 + cosθ + i sinθ.

Therefore, the modulus squared would be (1 + cosθ)^2 + (sinθ)^2.

Calculating that: (1 + 2cosθ + cos²θ) + sin²θ = 1 + 2cosθ + cos²θ + sin²θ = 1 + 2cosθ + 1 = 2 + 2cosθ.

So |1 + z| = sqrt(2 + 2cosθ) = sqrt(4cos²(θ/2)) = 2|cos(θ/2)|. Since θ is a real number, and cosine can be positive or negative, but since we're taking absolute value, it's 2|cos(θ/2)|.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
Now for the second modulus: |1 - z + z²|.

Let me compute this similarly. Let's write 1 - z + z² as 1 - e^{iθ} + e^{i2θ}.

<prompt_3.1> Compute the modulus |1 - z + z²| using trigonometric identities. </prompt_3.1>
-----

Based on execution_2.1:
Alternatively, maybe there is a smarter way to approach this problem without parameterizing z. Let me think.

<prompt_3.2> Consider alternative approaches to find the maximum value. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me compute the modulus squared:

|1 - e^{iθ} + e^{i2θ}|².

Let me denote this as |A|² where A = 1 - e^{iθ} + e^{i2θ}.

Expanding A:

First, write each term in terms of cos and sin:

1 = 1 + 0i

-e^{iθ} = -cosθ - i sinθ

e^{i2θ} = cos2θ + i sin2θ

Adding them together:

Real part: 1 - cosθ + cos2θ

Imaginary part: 0 - sinθ + sin2θ

So the modulus squared is [1 - cosθ + cos2θ]^2 + [ - sinθ + sin2θ]^2

Let me compute this step by step.
</execution_3.1>

<execution_3.2>
Since |z| = 1, we can use properties of complex numbers on the unit circle. For example, maybe using geometric interpretations or vector addition in the complex plane.

But perhaps the approach I took is standard for such problems. Let me see if there's a way to combine the terms or use triangle inequality or something else.

Alternatively, note that |1 - z + z²| can be rewritten. Let me see: 1 - z + z² is a quadratic in z. Maybe factor it? Let's check if it factors.

The roots of z² - z + 1 = 0 are [1 ± sqrt(1 - 4)]/2 = [1 ± i√3]/2. So the roots are complex numbers inside the unit circle since their modulus is sqrt(1/4 + 3/4) = 1. So maybe not helpful here.

Alternatively, perhaps express 1 - z + z² in terms of (1 + z + z²) - 2z. Wait, 1 + z + z² is a geometric series. But here we have 1 - z + z². Hmm. Alternatively, 1 - z + z² = (z³ + 1)/(z + 1) when z ≠ -1. Wait, let's check:

(z + 1)(z² - z + 1) = z³ + 1. So yes, if z ≠ -1, then 1 - z + z² = (z³ + 1)/(z + 1). But since |z| = 1, z ≠ -1 only when z ≠ -1. But even so, I don't know if this helps.

Alternatively, perhaps use the identity that |1 - z + z²| = |(z³ - 1)/(z - 1)| when z ≠ 1. But again, not sure.

Alternatively, maybe use the fact that |1 - z + z²| = |z² + (-z) + 1|. Since |z| = 1, |z²| = 1, so maybe interpret this as a triangle inequality problem.

But perhaps this is complicating things. The previous approach of parametrizing z as e^{iθ} seems straightforward and leads to an expression in terms of θ, which can then be maximized.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
First, compute the real part:

<prompt_4.1> Calculate the modulus |1 - z + z²| using trigonometric identities. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
1 - cosθ + cos2θ.

Recall that cos2θ = 2cos²θ - 1. So substituting:

1 - cosθ + 2cos²θ - 1 = 2cos²θ - cosθ.

Similarly, the imaginary part:

-sinθ + sin2θ.

Sin2θ = 2sinθ cosθ, so:

-sinθ + 2sinθ cosθ = sinθ (-1 + 2cosθ).

Therefore, modulus squared is [2cos²θ - cosθ]^2 + [sinθ (-1 + 2cosθ)]^2.

Factor out ( -1 + 2cosθ )² from both terms:

First term: [2cos²θ - cosθ]^2 = [cosθ (2cosθ - 1)]^2 = cos²θ (2cosθ - 1)^2

Second term: [sinθ (-1 + 2cosθ)]^2 = sin²θ (2cosθ - 1)^2

Therefore, modulus squared is [cos²θ + sin²θ] (2cosθ - 1)^2 = (1)(2cosθ - 1)^2.

Therefore, modulus is |2cosθ - 1|.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_3.1, execution_4.1:
Wait, that's interesting. So |1 - z + z²| = |2cosθ - 1|. Let me check if that's correct. Let me verify with a specific value.

<prompt_5.1> Verify the computed modulus |1 - z + z²|. </prompt_5.1>
-----

Based on execution_3.1, execution_3.2:
So going back, the original expression to maximize is |1 + z| + |1 - z + z²| = 2|cos(θ/2)| + |2cosθ - 1|.

So we need to find the maximum of 2|cos(θ/2)| + |2cosθ - 1| where θ is in [0, 2π). Let's denote x = cosθ. Then, since θ is in [0, 2π), x ∈ [-1, 1]. However, since we have θ/2, the variable substitution might not be straightforward.

<prompt_5.2> Solve the maximization problem using calculus and trigonometric identities. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Suppose θ = 0. Then z = 1. Then 1 - z + z² = 1 - 1 + 1 = 1. So modulus is 1. On the other hand, 2cos0 - 1 = 2*1 -1 =1. Absolute value is 1, which matches. 

Another test: θ = π/3. Then cosθ = 0.5. Then 2cosθ -1 = 0. Then modulus should be 0. Let's compute 1 - z + z² when θ = π/3. z = e^{iπ/3} = 0.5 + i(√3/2). Then z² = e^{i2π/3} = -0.5 + i(√3/2). So 1 - z + z² = 1 - (0.5 + i√3/2) + (-0.5 + i√3/2) = 1 -0.5 -0.5 + (-i√3/2 + i√3/2) = 0. So modulus is 0, which matches |2cosθ -1| = |2*(0.5) -1| = |1 -1| =0.
</execution_5.1>

<execution_5.2>
Alternatively, perhaps express everything in terms of cosθ.

But maybe it's better to use calculus here. Let's write the expression as a function of θ:

f(θ) = 2|cos(θ/2)| + |2cosθ - 1|.

Since cos(θ/2) is non-negative when θ ∈ [0, π], and negative when θ ∈ (π, 2π). But since it's absolute value, we can write:

For θ ∈ [0, π], |cos(θ/2)| = cos(θ/2)

For θ ∈ [π, 2π], |cos(θ/2)| = -cos(θ/2)

Similarly, 2cosθ - 1 is an expression that depends on θ. Let's find when 2cosθ - 1 is positive or negative.

2cosθ - 1 ≥ 0 ⇒ cosθ ≥ 1/2 ⇒ θ ∈ [0, π/3] ∪ [5π/3, 2π). Similarly, negative when θ ∈ (π/3, 5π/3).

Therefore, the expression f(θ) can be split into different intervals where the absolute value expressions change their sign.

So let's consider the intervals:

1. θ ∈ [0, π/3]
2. θ ∈ (π/3, π]
3. θ ∈ [π, 5π/3]
4. θ ∈ [5π/3, 2π)

In each interval, we can remove the absolute values accordingly.

Let's handle each interval:

Interval 1: θ ∈ [0, π/3]

Here, cos(θ/2) ≥ cos(π/6) = √3/2 > 0, so |cos(θ/2)| = cos(θ/2)

Also, 2cosθ - 1 ≥ 0 since θ ≤ π/3, so |2cosθ - 1| = 2cosθ - 1

Thus, f(θ) = 2cos(θ/2) + 2cosθ - 1

Interval 2: θ ∈ (π/3, π]

Here, cos(θ/2) is still positive since θ/2 ∈ (π/6, π/2), so |cos(θ/2)| = cos(θ/2)

But 2cosθ - 1 < 0 because θ > π/3, so |2cosθ - 1| = -(2cosθ - 1) = 1 - 2cosθ

Thus, f(θ) = 2cos(θ/2) + 1 - 2cosθ

Interval 3: θ ∈ [π, 5π/3]

Here, θ/2 ∈ [π/2, 5π/6], so cos(θ/2) ≤ 0, hence |cos(θ/2)| = -cos(θ/2)

Also, 2cosθ - 1: since θ ∈ [π, 5π/3], cosθ ∈ [-1, 1/2]. So 2cosθ - 1 ∈ [-3, 0], so |2cosθ - 1| = -(2cosθ - 1) = 1 - 2cosθ

Thus, f(θ) = -2cos(θ/2) + 1 - 2cosθ

Interval 4: θ ∈ [5π/3, 2π)

Here, θ/2 ∈ [5π/6, π), so cos(θ/2) ≤ 0, so |cos(θ/2)| = -cos(θ/2)

Also, 2cosθ -1: θ ∈ [5π/3, 2π), so cosθ ∈ (1/2, 1], so 2cosθ -1 ∈ (0, 1], so |2cosθ -1| = 2cosθ -1

Thus, f(θ) = -2cos(θ/2) + 2cosθ -1

So now, in each interval, we can express f(θ) without absolute values and then find its maximum by taking derivatives.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
Let's start with Interval 1: θ ∈ [0, π/3]

f(θ) = 2cos(θ/2) + 2cosθ -1

Let's compute its derivative:

<prompt_6.1> Find the maximum of the expression in Interval 1. </prompt_6.1>

Interval 2: θ ∈ (π/3, π]

f(θ) = 2cos(θ/2) +1 -2cosθ

Compute derivative:

<prompt_6.2> Find the maximum of the expression in Interval 2. </prompt_6.2>

Interval 3: θ ∈ [π, 5π/3]

f(θ) = -2cos(θ/2) +1 -2cosθ

Compute derivative:

<prompt_6.3> Find the maximum of the expression in Interval 3. </prompt_6.3>

Interval 4: θ ∈ [5π/3, 2π)

f(θ) = -2cos(θ/2) +2cosθ -1

Compute derivative:

<prompt_6.4> Find the maximum of the expression in Interval 4. </prompt_6.4>
</Plan_6>

<Execution_6>
<execution_6.1>
f’(θ) = -2*(1/2)sin(θ/2) -2sinθ = -sin(θ/2) -2sinθ

Set derivative to zero:

-sin(θ/2) -2sinθ = 0

sinθ/2 + 2sinθ = 0

But θ ∈ [0, π/3], so θ/2 ∈ [0, π/6], and sinθ and sinθ/2 are positive here. Therefore, the derivative is negative (since both terms are negative). Therefore, f(θ) is decreasing in this interval. Therefore, maximum occurs at θ = 0.

Compute f(0):

2cos(0) + 2cos0 -1 = 2*1 + 2*1 -1 = 2 + 2 -1 = 3

At θ = π/3:

2cos(π/6) + 2cos(π/3) -1 = 2*(√3/2) + 2*(1/2) -1 = √3 +1 -1 = √3 ≈1.732

So in interval 1, maximum is 3 at θ=0.
</execution_6.1>

<execution_6.2>
df/dθ = -sinθ + 2sinθ = sinθ(2 -1) = sinθ

Wait, wait: derivative of 2cos(θ/2) is -sin(θ/2) * (1/2) * 2 = -sin(θ/2). Wait, no:

Wait, derivative of cos(u) is -sin(u)*u', so derivative of 2cos(θ/2) is 2*(-sin(θ/2))*(1/2) = -sin(θ/2)

Similarly, derivative of 1 is 0, derivative of -2cosθ is 2sinθ

So total derivative: -sin(θ/2) + 2sinθ

Therefore, df/dθ = -sin(θ/2) + 2sinθ

Set derivative to zero:

-sin(θ/2) + 2sinθ = 0

Let me solve for θ:

2sinθ = sin(θ/2)

Express sinθ in terms of θ/2: sinθ = 2sin(θ/2)cos(θ/2)

Therefore:

2*(2sin(θ/2)cos(θ/2)) = sin(θ/2)

=> 4sin(θ/2)cos(θ/2) = sin(θ/2)

Assuming sin(θ/2) ≠ 0, we can divide both sides by sin(θ/2):

4cos(θ/2) = 1 ⇒ cos(θ/2) = 1/4 ⇒ θ/2 = arccos(1/4) ⇒ θ = 2arccos(1/4)

But θ is in (π/3, π]. Let's compute arccos(1/4):

arccos(1/4) is approximately 1.318 radians (since cos(1.318) ≈ 0.25). Then θ ≈ 2.636 radians. Which is approximately 150 degrees (π radians is 180 degrees). 2.636 radians is about 150.7 degrees, which is less than π (180 degrees). Wait, 2.636 radians is approximately 150.7 degrees, which is less than π (180 degrees). Wait, but θ is in (π/3, π], which is approximately (1.047, 3.142). So 2.636 is within this interval. So θ ≈ 2.636 radians is a critical point.

But we need to check if this is a maximum.

Alternatively, let's compute the second derivative or evaluate the function around this point. However, given the complexity, perhaps it's better to evaluate f(θ) at critical points and endpoints.

But let's note this critical point as θ1 = 2arccos(1/4). Let's compute f(θ1):

First, θ1 = 2arccos(1/4). Let me denote α = arccos(1/4), so θ1 = 2α. Then cosα = 1/4, so sinα = sqrt(1 - 1/16) = sqrt(15/16) = √15/4.

Compute f(θ1):

= 2cos(θ1/2) +1 -2cosθ1

But θ1/2 = α, so cos(θ1/2) = cosα = 1/4

cosθ1 = cos(2α) = 2cos²α -1 = 2*(1/16) -1 = 2/16 -1 = 1/8 -1 = -7/8

Thus, f(θ1) = 2*(1/4) +1 -2*(-7/8) = 0.5 +1 + 14/8 = 1.5 + 1.75 = 3.25

Wait, 2*(1/4) is 0.5, plus 1 is 1.5, then -2*(-7/8) is +14/8 = 1.75. So total 1.5 +1.75 = 3.25. So 3.25 is 13/4. Hmm.
</execution_6.2>

<execution_6.3>
f’(θ) = -2*(-sinθ/2)*(1/2) + 0 + 2sinθ

= sinθ/2 + 2sinθ

= sinθ/2 + 4sinθ/2

= (5/2)sinθ

Set derivative to zero:

(5/2)sinθ = 0 ⇒ sinθ = 0 ⇒ θ = 0, π

But in interval [π, 5π/3], θ = π is a critical point. Let's check endpoints θ = π and θ = 5π/3.

At θ = π:

f(π) = -2cos(π/2) +1 -2cosπ = -2*0 +1 -2*(-1) = 0 +1 +2 =3

At θ =5π/3:

f(5π/3) = -2cos(5π/6) +1 -2cos(5π/3)

cos(5π/6) = -√3/2, cos(5π/3) = 1/2

So:

-2*(-√3/2) +1 -2*(1/2) = √3 +1 -1 = √3 ≈1.732

So maximum at θ=π is 3.
</execution_6.3>

<execution_6.4>
So f’(θ) = -2*(-sin(θ/2))*(1/2) + 2*(-sinθ) = sin(θ/2) - 2sinθ

Set derivative to zero:

sin(θ/2) - 2sinθ =0

But θ ∈ [5π/3, 2π), so θ/2 ∈ [5π/6, π). Let me set φ = θ/2, so φ ∈ [5π/6, π). Then θ = 2φ, so equation becomes sinφ - 2sin2φ =0.

sinφ - 4sinφ cosφ =0

sinφ (1 -4cosφ)=0

Solutions when sinφ=0 or cosφ=1/4.

But φ ∈ [5π/6, π), so sinφ=0 only at φ=π. So φ=π is a solution.

For cosφ=1/4, φ= arccos(1/4). But arccos(1/4) ≈1.318 radians ≈75.5 degrees, which is less than 5π/6 ≈2.618 radians≈150 degrees. So no solution in this interval. Therefore, the only critical point is at φ=π, which is θ=2π. But θ=2π is the endpoint of the interval. So in this interval, the maximum occurs either at θ=5π/3 or θ=2π.

Compute f(5π/3):

θ=5π/3, so cosθ=cos(5π/3)=0.5, so 2cosθ -1=0. So f(θ)= -2cos(5π/6) +2cos(5π/3) -1.

cos(5π/6)= -√3/2, so -2*(-√3/2)=√3

cos(5π/3)=0.5, so 2*0.5=1. Thus, f(5π/3)=√3 +1 -1=√3≈1.732

At θ=2π, cosθ=1, so f(2π)= -2cos(π) +2cos2π -1= -2*(-1) +2*1 -1=2 +2 -1=3.

So in this interval, the maximum is 3 at θ=2π.

But θ=2π is also the endpoint of the previous interval.
</execution_6.4>
</Execution_6>

<Plan_7>
Based on execution_6.1, execution_6.2, execution_6.3, execution_6.4:
Wait, but θ=2π is the same as θ=0. So actually, we need to check all intervals and see where the maximum occurs.

So summarizing:

Interval 1: Maximum 3 at θ=0

Interval 2: Maximum at θ≈2.636 radians (θ1=2arccos(1/4)), giving f(θ)=3.25

Interval 3: Maximum 3 at θ=π

Interval 4: Maximum 3 at θ=2π (which is same as θ=0)

But wait, so the maximum value is 3.25? Which is 13/4.
But let's check if this is correct.

Wait, but when θ ≈2.636 radians, which is approximately 150 degrees, let's compute the actual value.

<prompt_7.1> Verify and confirm the maximum value found. </prompt_7.1>

But let's verify with θ=2arccos(1/4). Let me compute arccos(1/4):

<prompt_7.2> Re-evaluate the critical point and confirm the maximum value. </prompt_7.2>

But wait, let me check θ=2arccos(1/4). Let's compute θ:

<prompt_7.3> Re-confirm the maximum value through trigonometric identities. </prompt_7.3>

But is this the actual maximum?

Wait, but let's check another angle. Let me try θ=2π/3 (120 degrees).

<prompt_7.4> Test specific angles to verify the maximum value. </prompt_7.4>

Similarly, θ=π/2 (90 degrees):

<prompt_7.5> Further test specific angles to verify the maximum value. </prompt_7.5>

But let's check θ= arccos(1/4). Wait, θ=2arccos(1/4). Let's compute cosθ.

<prompt_7.6> Compute and verify the expression at the critical point. </prompt_7.6>

But wait, is this correct?

Alternatively, let's check θ=2π/3 (120 degrees). Then z=e^{i2π/3}= -1/2 +i√3/2.

<prompt_7.7> Re-evaluate the expression at θ=2π/3. </prompt_7.7>

But let me check another angle, say θ=π/3 (60 degrees):

<prompt_7.8> Test additional angles to ensure the maximum value is correct. </prompt_7.8>
</Plan_7>

<Execution_7>
<execution_7.1>
If θ ≈2.636 radians, which is 150 degrees. Let's compute f(θ):

First, θ=2.636≈2.636

θ/2≈1.318 radians≈75.5 degrees

cos(θ/2)=cos(arccos(1/4))=1/4.

cosθ=cos(2.636)=cos(150 degrees)= -√3/2≈-0.866.

Wait, wait, wait. If θ=2arccos(1/4), and arccos(1/4)≈1.318 radians, then θ≈2.636 radians, which is 150.7 degrees. Then cosθ=cos(2.636)=cos(π -0.506)= -cos(0.506)≈-0.875. But 2cosθ -1=2*(-0.875)-1= -1.75 -1= -2.75, and absolute value is 2.75. Wait, but in interval 2, we have θ ∈(π/3, π], which is up to π≈3.1416. So θ≈2.636 is within this interval.

But when we computed f(θ) at the critical point, we found 3.25. But is this the actual maximum?

Alternatively, perhaps we can check θ=2π/3 (120 degrees). Then θ=2π/3≈2.094 radians.

Compute f(θ):

θ=2π/3. Then θ/2=π/3. So cos(θ/2)=0.5. So 2|cos(θ/2)|=1.

2cosθ -1=2cos(2π/3)-1=2*(-0.5)-1=-1-1=-2. So |2cosθ -1|=2. Therefore, total f(θ)=1+2=3. Which is less than 3.25.

Alternatively, θ=arccos(1/4)*2≈2.636. Let's compute 2|cos(θ/2)| + |2cosθ -1|.

cos(θ/2)=1/4, so 2*(1/4)=0.5.

cosθ=cos(2.636)= approximately cos(150 degrees)= -√3/2≈-0.866. So 2cosθ -1≈2*(-0.866)-1≈-1.732 -1≈-2.732. Absolute value≈2.732. So total≈0.5 +2.732≈3.232, which is approximately 13/4=3.25. So that seems correct.

But is there a higher value?
</execution_7.1>

<execution_7.2>
arccos(1/4) ≈1.31811607 radians. Then θ≈2.63623214 radians.

Compute f(θ) at this point:

cos(θ/2) = cos(arccos(1/4)) =1/4

cosθ = cos(2arccos(1/4)) = 2cos²(arccos(1/4)) -1 = 2*(1/4)^2 -1 = 2*(1/16) -1 = 1/8 -1 = -7/8

Therefore, f(θ) = -2*(1/4) +1 -2*(-7/8) = -0.5 +1 +14/8 = 0.5 +1.75=2.25. Wait, that contradicts the earlier calculation. Wait, no:

Wait, in interval 2, the expression is f(θ)=2cos(θ/2) +1 -2cosθ

So substituting:

2*(1/4) +1 -2*(-7/8) = 0.5 +1 +14/8 = 1.5 +1.75=3.25. Yes, that's correct. 0.5 +1 is 1.5, 14/8 is 1.75, so total 3.25. So f(θ)=13/4=3.25. So that is higher than 3.
</execution_7.2>

<execution_7.3>
cosα=1/4, so α≈1.318 radians. Then θ≈2.636 radians. Let's compute cos(θ/2)=cosα=1/4, so cos(θ/2)=1/4. Then sin(θ/2)=sqrt(1 - (1/4)^2)=sqrt(15)/4≈0.9682.

Then, compute f(θ)=2cos(θ/2)+1 -2cosθ. So 2*(1/4)=0.5, plus 1 is 1.5. Then cosθ=cos(2α)=2cos²α -1=2*(1/16)-1=1/8 -1= -7/8. So -2cosθ= -2*(-7/8)=14/8=1.75. Therefore, total f(θ)=1.5 +1.75=3.25.
</execution_7.3>

<execution_7.4>
Then cosθ=cos(π/3)=0.5. Then compute |1 + z| + |1 - z + z²|.

z = e^{iπ/3} = 0.5 + i√3/2.

Compute 1 + z = 1.5 + i√3/2. Modulus squared: (1.5)^2 + (√3/2)^2 = 2.25 + 0.75 = 3. So modulus is √3 ≈1.732.

Compute 1 - z + z². z² = e^{i2π/3} = -0.5 + i√3/2.

So 1 - z + z² = 1 - (0.5 + i√3/2) + (-0.5 + i√3/2) = 1 -0.5 -0.5 + (-i√3/2 + i√3/2) = 0. So modulus is 0. Thus, total expression is √3 +0≈1.732.
</execution_7.4>

<execution_7.5>
Let's compute f(θ):

At θ=2π/3:

First, |1 + z|. z=e^{i2π/3}= -1/2 +i√3/2. So 1 + z =1 -1/2 +i√3/2=1/2 +i√3/2. Modulus is sqrt( (1/2)^2 + (√3/2)^2 )=sqrt(1/4 +3/4)=1. So |1 + z|=1.

|1 - z + z²|. Compute 1 - z + z². z²=e^{i4π/3}= -1/2 -i√3/2. So 1 - z + z²=1 - (-1/2 +i√3/2) + (-1/2 -i√3/2)=1 +1/2 -i√3/2 -1/2 -i√3/2=1 -i√3. Modulus is sqrt(1^2 + (√3)^2)=sqrt(1 +3)=2. So total expression is 1 +2=3.
</execution_7.5>

<execution_7.6>
If α=arccos(1/4), then θ=2α. Then cosθ=cos2α=2cos²α -1=2*(1/4)^2 -1=2*(1/16)-1=1/8 -1= -7/8. So θ= arccos(-7/8). Let's compute the expression f(θ)= -2cos(θ/2) +1 -2cosθ.

cos(θ/2)=cosα=1/4, cosθ= -7/8. So:

-2*(1/4) +1 -2*(-7/8)= -0.5 +1 +1.75= 3.25. So 13/4=3.25.
</execution_7.6>

<execution_7.7>
Compute |1 + z| = |1 + (-1/2 +i√3/2)| = |1/2 +i√3/2| = sqrt( (1/2)^2 + (√3/2)^2 ) = sqrt(1/4 +3/4)=1. So |1 + z|=1.

|1 - z + z²|: compute 1 - z + z². z² = e^{i4π/3}= -1/2 -i√3/2. So 1 - z + z² = 1 - (-1/2 +i√3/2) + (-1/2 -i√3/2) = 1 +1/2 -i√3/2 -1/2 -i√3/2 = 1 -i√3. So modulus is sqrt(1^2 + (√3)^2)=sqrt(1 +3)=2. So total expression is 1 +2=3. So at θ=π/3, the value is 3.
</execution_7.7>

<execution_7.8>
Then z=e^{iπ/3}=0.5 +i√3/2.

Compute |1 + z|=|1 +0.5 +i√3/2|=sqrt(1.5² + (√3/2)^2)=sqrt(2.25 +0.75)=sqrt(3)=≈1.732

Compute |1 - z + z²|:

1 - (0.5 +i√3/2) + (0.5 +i√3/2)^2

First, compute z²: (0.5 +i√3/2)^2=0.25 + 2*0.5*i√3/2 + (i√3/2)^2=0.25 +0.5i√3 - 3/4= -0.5 +0.5i√3

So 1 - z + z²=1 -0.5 -i√3/2 + (-0.5 +0.5i√3)= (1 -0.5 -0.5) + (-i√3/2 +0.5i√3)=0 +0=0. So |1 - z + z²|=0. So total expression is ≈1.732, which is less than 3.25.

Alternatively, take θ=2arccos(1/4). Let me compute cosθ:

Let’s denote α=arccos(1/4), so cosα=1/4. Then θ=2α. Then cosθ=cos2α=2cos²α -1=2*(1/4)^2 -1=2*(1/16)-1=1/8 -1= -7/8. So cosθ=-7/8. Then |cosθ/2|=|cosα|=1/4. Then |2cosθ -1|=|2*(-7/8)-1|=|-14/8 -8/8|=|-22/8|=11/4=2.75. So total f(θ)=0.5 +2.75=3.25.
</execution_7.8>
</Execution_7>

<Plan_8>
Based on execution_7.8:
But let me check another angle, say θ=π/2 (90 degrees):

<prompt_8.1> Finalize and confirm the maximum value of the expression. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
z=e^{iπ/2}=i.

Compute |1 + z|=|1 +i|=sqrt(1 +1)=sqrt(2)≈1.414.

Compute |1 - z + z²|=|1 -i + (-1)|=| -i | =1. So total expression≈1.414 +1≈2.414, which is less than 3.25.

Another angle, θ= arccos(1/4)≈75.5 degrees.

So θ≈1.318 radians.

Compute |1 + z|: 2cos(θ/2)=2*(1/4)=0.5.

|1 - z + z²|=2cosθ -1=2*(-7/8)-1= -14/8 -8/8= -22/8= -2.75. Absolute value 2.75. So total 0.5 +2.75=3.25.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
**Final Answer**
The maximum value is \boxed{\dfrac{13}{4}}.
</Plan_9>

<Execution_9>
No prompt to execute.
</Execution_9>

<Final_answer>
Therefore, the maximum value of the given expression is $\boxed{\frac{13}{4}}$.
</Final_answer>
</think>