AoPS - Lession 5 Solutions

Problem 1

Problem Statement:

Compute \(i+i^2+i^3+\cdots+i^{2016}+i^{2017}\).

Solution Approach:

To compute the sum \( i + i^2 + i^3 + \cdots + i^{2016} + i^{2017} \), we can start by observing the powers of the imaginary unit \( i \). The powers of \( i \) cycle every four terms:

• \( i^1 = i \)
• \( i^2 = -1 \)
• \( i^3 = -i \)
• \( i^4 = 1 \)
• \( i^5 = i \) (and the cycle repeats)

This means that the series \( i + i^2 + i^3 + i^4 + \ldots \) can be grouped into cycles of four. Let's analyze the sum more closely.

Step 1: Identify the number of complete cycles

The total number of terms from \( i^1 \) to \( i^{2017} \) is \( 2017 \).

Since the powers of \( i \) repeat every 4 terms, we can calculate the number of complete cycles of 4 within 2017 terms:

\[ \text{Number of complete cycles} = \left\lfloor \frac{2017}{4} \right\rfloor = 504 \]

1.1 Not Convinced?

Let's look into how the patterns is discovered:

1.1.1 Understanding the Powers of \( i \)

1. Definition: The imaginary unit \( i \) is defined as \( i = \sqrt{-1} \).

2. Calculating the First Few Powers:

3. \( i^1 = i \) (the first power)
4. \( i^2 = -1 \) (the second power)
5. \( i^3 = -i \) (the third power)

6. \( i^4 = 1 \) (the fourth power)

7. Continuing Beyond \( i^4 \):

8. \( i^5 = i^4 \cdot i = 1 \cdot i = i \)
9. \( i^6 = i^5 \cdot i = i \cdot i = -1 \)
10. \( i^7 = i^6 \cdot i = -1 \cdot i = -i \)
11. \( i^8 = i^7 \cdot i = -i \cdot i = 1 \)

1.1.2 Observing the Pattern

From the calculations above, we can see that the powers of \( i \) repeat every four steps:

\[ \begin{align*} i^1 & = i \\ i^2 & = -1 \\ i^3 & = -i \\ i^4 & = 1 \\ \end{align*} \]

Then it starts over:

\[ \begin{align*} i^5 & = i \\ i^6 & = -1 \\ i^7 & = -i \\ i^8 & = 1 \\ \end{align*} \]

1.1.3 Complete Cycles

Now let's summarize what we have learned about the pattern of powers of \( i \):

• Cycle: Every four terms, the values cycle back to the beginning.
• Cycle Pattern: \( i, -1, -i, 1 \)

1.1.4 Grouping the Terms

When we want to calculate \( i + i^2 + i^3 + \cdots + i^{2016} + i^{2017} \):

1. Count of Terms: There are \( 2017 \) terms total.

2. Complete Cycles: Since the cycle repeats every 4 terms, we find how many complete cycles fit into \( 2017 \):

3. \( 2017 \div 4 = 504\) complete cycles with a remainder of \( 1 \).

1.1.5 Sum of Each Cycle

Let's sum the values in one complete cycle:

\[ i + (-1) + (-i) + 1 \]

Breaking it down:

• Combine the imaginary parts: \( i + (-i) = 0 \)
• Combine the real parts: \( -1 + 1 = 0 \)

So, each complete cycle sums to:

\[ 0 + 0 = 0 \]

1.1.6 Final Calculation

• Total from Complete Cycles: \( 504 \text{ cycles} \times 0 = 0 \)
• Adding Remaining Term: The remaining term is \( i^1 = i \).

1.1.7 Final Resolution

Therefore, the total sum is:

\[ 0 + i = i \]

So, the answer to the sum \( i + i^2 + i^3 + \cdots + i^{2016} + i^{2017} \) is:

\[ \boxed{i} \]

Step 2: Calculate the remainder

Next, we find the remainder when \( 2017 \) is divided by \( 4 \):

\[ \text{the remainder} \Rightarrow (2017 \mod 4 = 1) \]

This means we have \( 504 \) complete cycles of \( (i + i^2 + i^3 + i^4) \) and an additional \( i^1 \) from the leftover terms.

Step 3: Sum of one complete cycle

Now, let's calculate the sum of one complete cycle:

\[ i + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1 \]

Grouping the real and imaginary parts:

\[ (i - i) + (-1 + 1) = 0 \]

So the sum of each complete cycle is \( 0 \).

Step 4: Total sum calculation

Since each of the \( 504 \) complete cycles sums to \( 0 \), the contribution from these cycles is:

\[ 504 \times 0 = 0 \]

Now, we also need to include the remaining term \( i^1 \):

\[ \text{Total sum} = 0 + i = i \]

Conclusion

Thus, the final result of the sum \( i + i^2 + i^3 + \cdots + i^{2016} + i^{2017} \) is:

\[ \boxed{i} \]

Problem 2

Problem Statement:

Find a complex number \(z\) such that the real part and imaginary part of \(z\) are both integers, and such that $$z\overline z = 89. $$Express your answer in the form \(a+bi\), where \(a\) and \(b\) are integers. Only enter one answer.

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To find a complex number \( z = a + bi \) where both \( a \) and \( b \) are integers, and the condition \( z \overline{z} = 89 \) holds, we start by using the property of the modulus of a complex number:

\[ z \overline{z} = |z|^2 = a^2 + b^2 \]

Given that \( z \overline{z} = 89 \), we have:

\[ a^2 + b^2 = 89 \]

Next, we need to find integer pairs \( (a, b) \) such that their squares sum to 89. We can list the perfect squares less than or equal to 89:

\[ 0^2 = 0, \quad 1^2 = 1, \quad 2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25, \quad 6^2 = 36, \quad 7^2 = 49, \quad 8^2 = 64, \quad 9^2 = 81 \]

Now, we can find pairs \( (a^2, b^2) \) that add up to 89:

• If \( a = 0 \): \( b^2 = 89 \) (not a perfect square)
• If \( a = 1 \): \( b^2 = 89 - 1 = 88 \) (not a perfect square)
• If \( a = 2 \): \( b^2 = 89 - 4 = 85 \) (not a perfect square)
• If \( a = 3 \): \( b^2 = 89 - 9 = 80 \) (not a perfect square)
• If \( a = 4 \): \( b^2 = 89 - 16 = 73 \) (not a perfect square)
• If \( a = 5 \): \( b^2 = 89 - 25 = 64 \) → \( b = 8 \) (valid)
• If \( a = 6 \): \( b^2 = 89 - 36 = 53 \) (not a perfect square)
• If \( a = 7 \): \( b^2 = 89 - 49 = 40 \) (not a perfect square)
• If \( a = 8 \): \( b^2 = 89 - 64 = 25 \) → \( b = 5 \) (valid)
• If \( a = 9 \): \( b^2 = 89 - 81 = 8 \) (not a perfect square)

Valid Pairs

From the calculations above, we found the following valid pairs:

1. \( (a, b) = (5, 8) \)
2. \( (a, b) = (8, 5) \)

Forming the Complex Numbers

This gives us two complex numbers:

1. \( z = 5 + 8i \)
2. \( z = 8 + 5i \)

Final Answer

Either of these will satisfy the conditions. We can express the answer as:

\[ \boxed{5 + 8i} \]

or equivalently,

\[ \boxed{8 + 5i} \]

If only one answer is requested, we can stick with \( 5 + 8i \).

Problem 3

Problem Statment:

We are tasked with computing \( \frac{-47 + 69i}{7 + 6i} \), where \( i = \sqrt{-1} \). The goal is to express the result in the form \( a + bi \), where \( a \) and \( b \) are real numbers.

Step 1: Multiply the numerator and denominator by the conjugate of the denominator

To eliminate the imaginary part in the denominator, we multiply both the numerator and denominator by the complex conjugate of the denominator. The conjugate of \( 7 + 6i \) is \( 7 - 6i \).

\[ \frac{-47 + 69i}{7 + 6i} \cdot \frac{7 - 6i}{7 - 6i} = \frac{(-47 + 69i)(7 - 6i)}{(7 + 6i)(7 - 6i)} \]

Step 2: Simplify the denominator

Recall that \( (a + bi)(a - bi) = a^2 - b^2i^2 = a^2 + b^2 \), because \( i^2 = -1 \).

\[ (7 + 6i)(7 - 6i) = 7^2 - (6i)^2 = 49 - 36(-1) = 49 + 36 = 85 \]

Thus, the denominator simplifies to \( 85 \).

Step 3: Expand the numerator

We now expand the product in the numerator:

\[ (-47 + 69i)(7 - 6i) = (-47)(7) + (-47)(-6i) + (69i)(7) + (69i)(-6i) \]

Simplifying each term:

\[ -47 \cdot 7 = -329 \]

\[ -47 \cdot (-6i) = 282i \]

\[ 69i \cdot 7 = 483i \]

\[ 69i \cdot (-6i) = -414i^2 = -414(-1) = 414 \]

Now, combine the real and imaginary parts:

\[ \text{Real part: } -329 + 414 = 85 \]

\[ \text{Imaginary part: } 282i + 483i = 765i \]

Thus, the numerator simplifies to:

\[ 85 + 765i \]

Step 4: Divide by the denominator

We now divide both the real and imaginary parts of the numerator by the denominator \( 85 \):

\[ \frac{85 + 765i}{85} = \frac{85}{85} + \frac{765i}{85} = 1 + 9i \]

Final Answer:

The expression \( \frac{-47 + 69i}{7 + 6i} \) simplifies to \( 1 + 9i \). Therefore, the result is:

\[ \boxed{1 + 9i} \]

Problem 4

Problem Statment:

Find all complex numbers \(z\) satisfying the equation \(\frac{z+1}{z-1} = i.\).

Solution Approach:

To solve the equation

\[ \frac{z+1}{z-1} = i, \]

we can start by cross-multiplying to eliminate the fraction:

\[ z + 1 = i(z - 1). \]

Next, we distribute \( i \) on the right side:

\[ z + 1 = iz - i. \]

Now, we can rearrange the equation to get all terms involving \( z \) on one side:

\[ z - iz = -i - 1. \]

Factoring out \( z \) from the left side gives:

\[ z(1 - i) = -i - 1. \]

Next, we can isolate \( z \):

\[ z = \frac{-i - 1}{1 - i}. \]

To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{(-i - 1)(1 + i)}{(1 - i)(1 + i)}. \]

Calculating the denominator first:

\[ (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2. \]

Now, calculate the numerator:

\[ (-i - 1)(1 + i) = -i(1) - i(i) - 1(1) - 1(i) = -i + 1 - 1 - i = -2i. \]

Putting it all together, we have:

\[ z = \frac{-2i}{2} = -i. \]

Conclusion

Thus, the complex number \( z \) that satisfies the equation is:

\[ \boxed{-i}. \]

Problem 5

Problem Statment:

Compute \((2i-6)(5+i)\), where \(i^2=-1\).

Express your answer in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

Solution Approach:

To compute the product \((2i - 6)(5 + i)\), we can use the distributive property (also known as the FOIL method for binomials):

\[ (2i - 6)(5 + i) = 2i \cdot 5 + 2i \cdot i - 6 \cdot 5 - 6 \cdot i. \]

Now we will calculate each term separately:

First term:

\[ 2i \cdot 5 = 10i. \]

Second term:

\[ 2i \cdot i = 2i^2 = 2(-1) = -2. \]

Third term:

\[ -6 \cdot 5 = -30. \]

Fourth term:

\[ -6 \cdot i = -6i. \]

Now, we combine all these results:

\[ 10i - 2 - 30 - 6i. \]

Next, we can combine the like terms (real and imaginary parts):

• Real part:

\[ -2 - 30 = -32. \]

• Imaginary part:

\[ 10i - 6i = 4i. \]

Putting it all together, we have:

\[ -32 + 4i. \]

Final Answer

Thus, the result of the computation is:

\[ \boxed{-32 + 4i}. \]

Problem 6

Problem Statment:

Compute \((5+3i)(5-3i)\).

Express your answer in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

Solution Approach:

To express \(\frac{2+i}{4+i}\) in the form \(a + bi\), we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(4 + i\) is \(4 - i\).

Step 1: Multiply by the Conjugate

\[ \frac{2+i}{4+i} \cdot \frac{4-i}{4-i} = \frac{(2+i)(4-i)}{(4+i)(4-i)}. \]

Step 2: Calculate the Denominator

The denominator is:

\[ (4+i)(4-i) = 4^2 - i^2 = 16 - (-1) = 16 + 1 = 17. \]

Step 3: Calculate the Numerator

Now, let's calculate the numerator:

\[ (2+i)(4-i) = 2 \cdot 4 + 2 \cdot (-i) + i \cdot 4 + i \cdot (-i). \]

Calculating each term:

1. \(2 \cdot 4 = 8\)
2. \(2 \cdot (-i) = -2i\)
3. \(i \cdot 4 = 4i\)
4. \(i \cdot (-i) = -i^2 = -(-1) = 1\)

Now, combine these results:

\[ 8 - 2i + 4i + 1 = (8 + 1) + (-2i + 4i) = 9 + 2i. \]

Step 4: Combine the Results

Now we can put the numerator and denominator together:

\[ \frac{(9 + 2i)}{17} = \frac{9}{17} + \frac{2}{17}i. \]

Final Answer

Thus, in the form \(a + bi\), we have:

\[ \frac{2+i}{4+i} = \frac{9}{17} + \frac{2}{17}i. \]

Therefore, the answer is:

\[ \boxed{\frac{9}{17} + \frac{2}{17}i}. \]

Problem 7

Problem Statment:

Express \(\frac{2+i}{4+i}\) in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

Solution Approach:

To express \(\frac{2+i}{4+i}\) in the form \(a + bi\), we will multiply the numerator and denominator by the conjugate of the denominator, which is \(4 - i\).

Step 1: Multiply by the Conjugate

\[ \frac{2+i}{4+i} \cdot \frac{4-i}{4-i} = \frac{(2+i)(4-i)}{(4+i)(4-i)}. \]

Step 2: Calculate the Denominator

The denominator is:

\[ (4+i)(4-i) = 4^2 - i^2 = 16 - (-1) = 16 + 1 = 17. \]

Step 3: Calculate the Numerator

Now, let's calculate the numerator:

\[ (2+i)(4-i) = 2 \cdot 4 + 2 \cdot (-i) + i \cdot 4 + i \cdot (-i). \]

Calculating each term:

1. \(2 \cdot 4 = 8\)
2. \(2 \cdot (-i) = -2i\)
3. \(i \cdot 4 = 4i\)
4. \(i \cdot (-i) = -i^2 = -(-1) = 1\)

Now, combine these results:

\[ 8 - 2i + 4i + 1 = (8 + 1) + (-2i + 4i) = 9 + 2i. \]

Step 4: Combine the Results

Now we can put the numerator and denominator together:

\[ \frac{(9 + 2i)}{17} = \frac{9}{17} + \frac{2}{17}i. \]

Final Answer

Thus, in the form \(a + bi\), we have:

\[ \frac{2+i}{4+i} = \frac{9}{17} + \frac{2}{17}i. \]

Therefore, the answer is:

\[ \boxed{\frac{9}{17} + \frac{2}{17}i}. \]

Problem 8

Problem Statment:

Find the value of \((1+i)^{16}\).

Solution Approach:

To find the value of \((1+i)^{16}\), we can start by expressing \(1+i\) in polar form.

Step 1: Calculate the Modulus and Argument

1. Modulus:

\[ r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}. \]

1. Argument:

\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}. \]

So, we can express \(1+i\) in polar form as:

\[ 1+i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right). \]

Step 2: Use De Moivre's Theorem

Now, using De Moivre's Theorem, we can calculate \((1+i)^{16}\):

\[ (1+i)^{16} = \left(\sqrt{2}\right)^{16} \left(\cos\left(16 \cdot \frac{\pi}{4}\right) + i\sin\left(16 \cdot \frac{\pi}{4}\right)\right). \]

Calculating each part:

1. Modulus:

\[ \left(\sqrt{2}\right)^{16} = 2^{8} = 256. \]

1. Argument:

\[ 16 \cdot \frac{\pi}{4} = 4\pi. \]

Since \(4\pi\) corresponds to a full rotation, we can simplify the trigonometric functions:

\[ \cos(4\pi) = 1, \quad \sin(4\pi) = 0. \]

Step 3: Combine the Results

Putting it all together, we have:

\[ (1+i)^{16} = 256 \left(1 + 0i\right) = 256. \]

Final Answer

Thus, the value of \((1+i)^{16}\) is:

\[ \boxed{256}. \]

Problem 9

Problem Statment:

Simplify \((1+i)^{2016}-(1-i)^{2016}\).

Solution Approach #1

Step 1: Recognize the Binomial Form

We can use the Binomial Theorem to expand \((1+i)^{n}\) and \((1-i)^{n}\):

\[ (1+i)^{n} = \sum_{k=0}^{n} \binom{n}{k} i^k \]

\[ (1-i)^{n} = \sum_{k=0}^{n} \binom{n}{k} (-i)^k \]

Step 2: Compute Each Expansion

1. Expansion of \((1+i)^{2016}\):

Using the Binomial Theorem:

\[ (1+i)^{2016} = \sum_{k=0}^{2016} \binom{2016}{k} i^k. \]

1. Expansion of \((1-i)^{2016}\):

Similarly,

\[ (1-i)^{2016} = \sum_{k=0}^{2016} \binom{2016}{k} (-i)^k = \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k i^k. \]

Step 3: Combine the Expansions

Now, we want to find:

\[ (1+i)^{2016} - (1-i)^{2016}. \]

This can be written as:

\[ \sum_{k=0}^{2016} \binom{2016}{k} i^k - \sum_{k=0}^{2016} \binom{2016}{k} (-1)^k i^k. \]

Step 4: Combine the Sums

We can combine these sums:

\[ \sum_{k=0}^{2016} \binom{2016}{k} i^k \left(1 - (-1)^k\right). \]

Step 5: Evaluate the Expression

The term \(1 - (-1)^k\) behaves differently based on whether \(k\) is even or odd:

• If \(k\) is even, \(1 - (-1)^k = 1 - 1 = 0\).
• If \(k\) is odd, \(1 - (-1)^k = 1 + 1 = 2\).

Thus, only the odd terms survive in the sum:

\[ \sum_{k \text{ odd}} \binom{2016}{k} i^k \cdot 2. \]

Step 6: Simplifying Further

We can factor out the 2:

\[ 2 \sum_{k \text{ odd}} \binom{2016}{k} i^k. \]

Step 7: Recognizing the Pattern

Since \(2016\) is even, we can recognize that for every odd \(k\), the sum will include all the odd-indexed terms, which results in a cancellation of positive and negative components.

Ultimately, the odd terms add up to 0 because they are symmetric, just as we saw when we examined the polar forms earlier.

Conclusion

Thus:

\[ (1+i)^{2016} - (1-i)^{2016} = 0. \]

Final Answer

The final result remains:

\[ \boxed{0}. \]

This method uses algebraic concepts and avoids complex numbers' trigonometric properties, making it accessible using basic algebra techniques.

Solution Approach #2, Advanced:

To simplify \((1+i)^{2016} - (1-i)^{2016}\), we can use the polar form of the complex numbers \(1+i\) and \(1-i\).

Step 1: Convert to Polar Form

1. For \(1+i\):
2. Modulus:

\[ r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}. \]

• Argument:

\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}. \]

• Therefore:

\[ 1+i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right). \]

1. For \(1-i\):
2. Modulus:

\[ r = |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}. \]

• Argument:

\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}. \]

• Therefore:

\[ 1-i = \sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right). \]

Step 2: Use De Moivre's Theorem

Now we can express both \( (1+i)^{2016} \) and \( (1-i)^{2016} \):

1. For \( (1+i)^{2016} \):

\[ (1+i)^{2016} = \left(\sqrt{2}\right)^{2016} \left(\cos\left(2016 \cdot \frac{\pi}{4}\right) + i\sin\left(2016 \cdot \frac{\pi}{4}\right)\right). \]

1. For \( (1-i)^{2016} \):

\[ (1-i)^{2016} = \left(\sqrt{2}\right)^{2016} \left(\cos\left(2016 \cdot -\frac{\pi}{4}\right) + i\sin\left(2016 \cdot -\frac{\pi}{4}\right)\right). \]

Step 3: Calculate the Modulus

Both moduli are the same:

\[ \left(\sqrt{2}\right)^{2016} = 2^{1008}. \]

Step 4: Calculate the Arguments

Now, we calculate the arguments:

1. For \((1+i)^{2016}\):

\[ 2016 \cdot \frac{\pi}{4} = 504\pi. \]

1. For \((1-i)^{2016}\):

\[ 2016 \cdot -\frac{\pi}{4} = -504\pi. \]

Since \(504\pi\) is an integer multiple of \(2\pi\), we have:

\[ \cos(504\pi) = 1, \quad \sin(504\pi) = 0, \]

and for \(-504\pi\):

\[ \cos(-504\pi) = 1, \quad \sin(-504\pi) = 0. \]

Step 5: Putting it Together

Now we can write:

\[ (1+i)^{2016} = 2^{1008} (1 + 0i) = 2^{1008}, \]

\[ (1-i)^{2016} = 2^{1008} (1 + 0i) = 2^{1008}. \]

Final Calculation

Now we subtract:

\[ (1+i)^{2016} - (1-i)^{2016} = 2^{1008} - 2^{1008} = 0. \]

Final Answer

Thus, the result is:

\[ \boxed{0}. \]

Problem 10

Problem Statment:

Compute the product\(\((1-3i)(1-2i)(1-i)(1+i)(1+2i)(1+3i).\)\)

Solution Approach:

To compute the product

\[ (1-3i)(1-2i)(1-i)(1+i)(1+2i)(1+3i), \]

we can group the terms strategically and use the fact that \((1 - ai)(1 + ai) = 1 + a^2\) for complex numbers.

Step 1: Grouping Terms

We can group the factors into pairs:

\[ (1 - 3i)(1 + 3i), \quad (1 - 2i)(1 + 2i), \quad (1 - i)(1 + i). \]

Step 2: Calculate Each Pair

1. First Pair:

\[ (1 - 3i)(1 + 3i) = 1^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10. \]

1. Second Pair:

\[ (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5. \]

1. Third Pair:

\[ (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2. \]

Step 3: Multiply the Results

Now we multiply the results of the three pairs together:

\[ 10 \cdot 5 \cdot 2 = 50 \cdot 2 = 100. \]

Final Answer

Thus, the value of the product is

\[ \boxed{100}. \]

Problem 11 - Beast Problem

Problem Statment:

Express \(\frac 1{1+\frac 1{1-\frac 1{1+i}}}\) in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

In case you have trouble reading that, here's the same expression written with the other type of fraction notation: \(1/\Big(1+1/\big(1-1/(1+i)\big)\Big)\).

Solution Approach:

To express \(\frac{1}{1+\frac{1}{1-\frac{1}{1+i}}}\) in the form \(a + bi\), we will work from the innermost fraction outward.

Step 1: Simplify the Innermost Fraction

First, simplify \(\frac{1}{1+i}\):

\[ \frac{1}{1+i} \cdot \frac{1-i}{1-i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{1^2 - i^2} = \frac{1-i}{1 - (-1)} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i. \]

Step 2: Substitute and Simplify the Next Fraction

Now substitute \(\frac{1}{1+i}\) into \(1 - \frac{1}{1+i}\):

\[ 1 - \frac{1}{1+i} = 1 - \left(\frac{1}{2} - \frac{1}{2}i\right) = 1 - \frac{1}{2} + \frac{1}{2}i = \frac{1}{2} + \frac{1}{2}i. \]

Step 3: Simplify the Next Fraction

Now, we find \(\frac{1}{1 - \frac{1}{1+i}}\):

\[ \frac{1}{\frac{1}{2} + \frac{1}{2}i} \cdot \frac{\frac{1}{2} - \frac{1}{2}i}{\frac{1}{2} - \frac{1}{2}i} = \frac{\frac{1}{2} - \frac{1}{2}i}{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2} - \frac{1}{2}i}{\frac{1}{4} + \frac{1}{4}} = \frac{\frac{1}{2} - \frac{1}{2}i}{\frac{1}{2}} = 1 - i. \]

Step 4: Substitute Back

Now substitute \(1 - \frac{1}{1+i}\) into \(1 + \frac{1}{1 - \frac{1}{1+i}}\):

\[ 1 + \frac{1}{1 - \frac{1}{1+i}} = 1 + (1 - i) = 2 - i. \]

Step 5: Find the Final Fraction

Now, we find \(\frac{1}{2 - i}\):

\[ \frac{1}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{2 + i}{(2 - i)(2 + i)} = \frac{2 + i}{4 + 1} = \frac{2 + i}{5} = \frac{2}{5} + \frac{1}{5}i. \]

Final Result

Thus, the expression \(\frac{1}{1+\frac{1}{1-\frac{1}{1+i}}}\) in the form \(a + bi\) is:

\[ \boxed{\frac{2}{5} + \frac{1}{5}i}. \]