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Question 1
A complex number $z$ lies on the unit circle, so $∣ z ∣ = 1$ and $z = cos θ + i sin θ$ . Which of the following correctly expresses $z + \frac{1}{z}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 cos θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recognise the structure
We have $z = cos θ + i sin θ$ on the unit circle. We need to compute $z + \frac{1}{z}$ .
Step 2: Find $\frac{1}{z}$ using De Moivre's theorem
Since $∣ z ∣ = 1$ , we have $\frac{1}{z} = z^{- 1} = cos (- θ) + i sin (- θ) = cos θ - i sin θ$ .
Step 3: Add $z$ and $\frac{1}{z}$
$z + \frac{1}{z} = (cos θ + i sin θ) + (cos θ - i sin θ) = 2 cos θ$
Step 4: State the result
The imaginary parts cancel and we are left with $2 cos θ$ . This is a standard result used when deriving multiple-angle identities.
Method #2Approach 2
Step 1: Identify what is being tested
We need $z + \frac{1}{z}$ where $z$ is on the unit circle. We use the fact that $\frac{1}{z} = \overset{z}{ˉ} = cos θ - i sin θ$ when $∣ z ∣ = 1$ .
Step 2: Eliminate '$2i\sin\theta$'
$z - \frac{1}{z}$ gives $2 i sin θ$ , not $z + \frac{1}{z}$ . This option confuses addition with subtraction.
Step 3: Eliminate '$2\cos\theta + 2i\sin\theta$'
This equals $2 z$ , not $z + \frac{1}{z}$ . It arises from incorrectly assuming $\frac{1}{z} = z$ .
Step 4: Eliminate '$\cos(2\theta) + i\sin(2\theta)$'
This equals $z^{2}$ by De Moivre's theorem, not $z + \frac{1}{z}$ . Confusing multiplication with addition is a common error.
Step 5: Select the correct answer
Adding $z$ and $\overset{z}{ˉ}$ gives twice the real part: $2 cos θ$ . The correct answer is $2 cos θ$ .
Question 2
Let $z = cos θ + i sin θ$ with $∣ z ∣ = 1$ . Which expression is equivalent to $z^{n} + z^{- n}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 cos (n θ)$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Apply De Moivre's theorem to $z^n$
By De Moivre's theorem, $z^{n} = cos (n θ) + i sin (n θ)$ and $z^{- n} = cos (- n θ) + i sin (- n θ) = cos (n θ) - i sin (n θ)$ .
Step 2: Add $z^n$ and $z^{-n}$
$z^{n} + z^{- n} = (cos (n θ) + i sin (n θ)) + (cos (n θ) - i sin (n θ)) = 2 cos (n θ)$
Step 3: Confirm the result
The imaginary parts cancel, yielding $2 cos (n θ)$ . This generalises the $n = 1$ case and is used to derive $cos (n θ)$ expansions.
Method #2Approach 2
Step 1: Identify the key theorem needed
De Moivre's theorem states $(cos θ + i sin θ)^{n} = cos (n θ) + i sin (n θ)$ . We need $z^{n} + z^{- n}$ .
Step 2: Eliminate '$2i\sin(n\theta)$'
This equals $z^{n} - z^{- n}$ , not $z^{n} + z^{- n}$ . It results from subtracting rather than adding the two conjugates.
Step 3: Eliminate '$2\cos\theta$'
This is the result for $n = 1$ only. The general expression must involve $n θ$ , not just $θ$ .
Step 4: Eliminate '$\cos(n\theta)$'
This is missing the factor of 2. Summing two equal real parts gives $2 cos (n θ)$ , not $cos (n θ)$ .
Step 5: Select the correct answer
The correct answer is $2 cos (n θ)$ , obtained by adding $z^{n}$ and its conjugate $z^{- n}$ .

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Free preview 2/16

Question 1
A complex number $z$ lies on the unit circle, so $∣ z ∣ = 1$ and $z = cos θ + i sin θ$ . Which of the following correctly expresses $z + \frac{1}{z}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 cos θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recognise the structure
We have $z = cos θ + i sin θ$ on the unit circle. We need to compute $z + \frac{1}{z}$ .
Step 2: Find $\frac{1}{z}$ using De Moivre's theorem
Since $∣ z ∣ = 1$ , we have $\frac{1}{z} = z^{- 1} = cos (- θ) + i sin (- θ) = cos θ - i sin θ$ .
Step 3: Add $z$ and $\frac{1}{z}$
$z + \frac{1}{z} = (cos θ + i sin θ) + (cos θ - i sin θ) = 2 cos θ$
Step 4: State the result
The imaginary parts cancel and we are left with $2 cos θ$ . This is a standard result used when deriving multiple-angle identities.
Method #2Approach 2
Step 1: Identify what is being tested
We need $z + \frac{1}{z}$ where $z$ is on the unit circle. We use the fact that $\frac{1}{z} = \overset{z}{ˉ} = cos θ - i sin θ$ when $∣ z ∣ = 1$ .
Step 2: Eliminate '$2i\sin\theta$'
$z - \frac{1}{z}$ gives $2 i sin θ$ , not $z + \frac{1}{z}$ . This option confuses addition with subtraction.
Step 3: Eliminate '$2\cos\theta + 2i\sin\theta$'
This equals $2 z$ , not $z + \frac{1}{z}$ . It arises from incorrectly assuming $\frac{1}{z} = z$ .
Step 4: Eliminate '$\cos(2\theta) + i\sin(2\theta)$'
This equals $z^{2}$ by De Moivre's theorem, not $z + \frac{1}{z}$ . Confusing multiplication with addition is a common error.
Step 5: Select the correct answer
Adding $z$ and $\overset{z}{ˉ}$ gives twice the real part: $2 cos θ$ . The correct answer is $2 cos θ$ .
Question 2
Let $z = cos θ + i sin θ$ with $∣ z ∣ = 1$ . Which expression is equivalent to $z^{n} + z^{- n}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 cos (n θ)$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Apply De Moivre's theorem to $z^n$
By De Moivre's theorem, $z^{n} = cos (n θ) + i sin (n θ)$ and $z^{- n} = cos (- n θ) + i sin (- n θ) = cos (n θ) - i sin (n θ)$ .
Step 2: Add $z^n$ and $z^{-n}$
$z^{n} + z^{- n} = (cos (n θ) + i sin (n θ)) + (cos (n θ) - i sin (n θ)) = 2 cos (n θ)$
Step 3: Confirm the result
The imaginary parts cancel, yielding $2 cos (n θ)$ . This generalises the $n = 1$ case and is used to derive $cos (n θ)$ expansions.
Method #2Approach 2
Step 1: Identify the key theorem needed
De Moivre's theorem states $(cos θ + i sin θ)^{n} = cos (n θ) + i sin (n θ)$ . We need $z^{n} + z^{- n}$ .
Step 2: Eliminate '$2i\sin(n\theta)$'
This equals $z^{n} - z^{- n}$ , not $z^{n} + z^{- n}$ . It results from subtracting rather than adding the two conjugates.
Step 3: Eliminate '$2\cos\theta$'
This is the result for $n = 1$ only. The general expression must involve $n θ$ , not just $θ$ .
Step 4: Eliminate '$\cos(n\theta)$'
This is missing the factor of 2. Summing two equal real parts gives $2 cos (n θ)$ , not $cos (n θ)$ .
Step 5: Select the correct answer
The correct answer is $2 cos (n θ)$ , obtained by adding $z^{n}$ and its conjugate $z^{- n}$ .

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Step-by-step walkthrough

Step 1: Recognise the structure

Step 2: Find $\frac{1}{z}$ using De Moivre's theorem

Step 3: Add $z$ and $\frac{1}{z}$

Step 4: State the result

Step 1: Identify what is being tested

Step 2: Eliminate '$2i\sin\theta$'

Step 3: Eliminate '$2\cos\theta + 2i\sin\theta$'

Step 4: Eliminate '$\cos(2\theta) + i\sin(2\theta)$'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Apply De Moivre's theorem to $z^n$

Step 2: Add $z^n$ and $z^{-n}$

Step 3: Confirm the result

Step 1: Identify the key theorem needed

Step 2: Eliminate '$2i\sin(n\theta)$'

Step 3: Eliminate '$2\cos\theta$'

Step 4: Eliminate '$\cos(n\theta)$'

Step 5: Select the correct answer

14 more questions in this topic

Step-by-step walkthrough

Step 1: Recognise the structure

Step 2: Find $\frac{1}{z}$ using De Moivre's theorem

Step 3: Add $z$ and $\frac{1}{z}$

Step 4: State the result

Step 1: Identify what is being tested

Step 2: Eliminate '$2i\sin\theta$'

Step 3: Eliminate '$2\cos\theta + 2i\sin\theta$'

Step 4: Eliminate '$\cos(2\theta) + i\sin(2\theta)$'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Apply De Moivre's theorem to $z^n$

Step 2: Add $z^n$ and $z^{-n}$

Step 3: Confirm the result

Step 1: Identify the key theorem needed

Step 2: Eliminate '$2i\sin(n\theta)$'

Step 3: Eliminate '$2\cos\theta$'

Step 4: Eliminate '$\cos(n\theta)$'

Step 5: Select the correct answer

14 more questions in this topic