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Question 1
Let $w = cos θ + i sin θ$ where $0 < θ < π$ . Which of the following correctly expresses $\frac{1}{w}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $cos θ - i sin θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recognise the form of $w$
Since $w = cos θ + i sin θ$ , we have $∣ w ∣ = 1$ and $ar g (w) = θ$ . In Euler form, $w = e^{i θ}$ .
Step 2: Compute the reciprocal
$\frac{1}{w} = \frac{1}{e ^{i θ}} = e^{- i θ} = cos (- θ) + i sin (- θ)$
Step 3: Use even/odd identities
Since $cos (- θ) = cos θ$ and $sin (- θ) = - sin θ$ , we get $\frac{1}{w} = cos θ - i sin θ$ .
Step 4: Confirm the answer
This is simply the complex conjugate $\overset{w}{ˉ}$ , reflecting the fact that for $∣ w ∣ = 1$ , $w^{- 1} = \overset{w}{ˉ}$ . The correct answer is $cos θ - i sin θ$ .
Method #2Approach 2
Step 1: Identify what is being asked
We need the multiplicative inverse of $w = cos θ + i sin θ$ , which must satisfy $w \cdot \frac{1}{w} = 1$ .
Step 2: Eliminate $\cos\theta + i\sin\theta$
If $\frac{1}{w} = cos θ + i sin θ = w$ , then $w^{2} = 1$ , meaning $θ = 0$ or $θ = π$ . But $0 < θ < π$ , so this only works at boundary values — eliminate.
Step 3: Eliminate $-\cos\theta + i\sin\theta$
Multiplying $w \cdot (- cos θ + i sin θ) = - cos^{2} θ + i sin θ cos θ + i sin θ cos θ - sin^{2} θ = - (cos^{2} θ + sin^{2} θ) + 2 i sin θ cos θ = - 1 + i sin 2 θ \neq = 1$ . Eliminate.
Step 4: Eliminate $-\cos\theta - i\sin\theta$
This equals $- w$ , so $w \cdot (- w) = - w^{2} = - (cos 2 θ + i sin 2 θ) \neq = 1$ in general. Eliminate.
Step 5: Select $\cos\theta - i\sin\theta$
Checking: $(cos θ + i sin θ) (cos θ - i sin θ) = cos^{2} θ + sin^{2} θ = 1$ ✓. This is the correct answer.
Question 2
Let $w = cos θ + i sin θ$ where $0 < θ < π$ . What is $ar g (w^{2})$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recall the multiplication rule for arguments
For any two complex numbers, $ar g (z_{1} z_{2}) = ar g (z_{1}) + ar g (z_{2})$ . Applying this to $w^{2} = w \cdot w$ :
Step 2: Apply the argument addition rule
$ar g (w^{2}) = ar g (w) + ar g (w) = 2 ar g (w) = 2 θ$
Step 3: Verify using Euler form
Since $w = e^{i θ}$ , we have $w^{2} = e^{2 i θ} = cos 2 θ + i sin 2 θ$ , so $ar g (w^{2}) = 2 θ$ .
Step 4: Confirm
The argument of $w^{2}$ is $2 θ$ , consistent with the general rule that squaring doubles the argument.
Method #2Approach 2
Step 1: What is being asked
We need the argument of $w^{2}$ given $w = e^{i θ}$ with $0 < θ < π$ .
Step 2: Eliminate $\theta^2$
The argument is an angle, not a product of $θ$ with itself. Arguments combine by addition under multiplication, not by multiplying the arguments together. Eliminate $θ^{2}$ .
Step 3: Eliminate $\dfrac{\theta}{2}$
$\frac{θ}{2}$ would be the argument of $w^{1/2}$ (a square root of $w$ ), not of $w^{2}$ . Squaring multiplies the exponent, not halves it. Eliminate.
Step 4: Eliminate $\theta + \pi$
Adding $π$ to the argument corresponds to multiplying by $e^{iπ} = - 1$ , i.e., negating $w$ . This is not the same as squaring $w$ . Eliminate.
Step 5: Select $2\theta$
Squaring doubles the argument: $ar g (w^{2}) = 2 θ$ . This follows directly from $e^{i θ} \cdot e^{i θ} = e^{2 i θ}$ .

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13 more questions in this topic

Free preview 2/15

Question 1
Let $w = cos θ + i sin θ$ where $0 < θ < π$ . Which of the following correctly expresses $\frac{1}{w}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $cos θ - i sin θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recognise the form of $w$
Since $w = cos θ + i sin θ$ , we have $∣ w ∣ = 1$ and $ar g (w) = θ$ . In Euler form, $w = e^{i θ}$ .
Step 2: Compute the reciprocal
$\frac{1}{w} = \frac{1}{e ^{i θ}} = e^{- i θ} = cos (- θ) + i sin (- θ)$
Step 3: Use even/odd identities
Since $cos (- θ) = cos θ$ and $sin (- θ) = - sin θ$ , we get $\frac{1}{w} = cos θ - i sin θ$ .
Step 4: Confirm the answer
This is simply the complex conjugate $\overset{w}{ˉ}$ , reflecting the fact that for $∣ w ∣ = 1$ , $w^{- 1} = \overset{w}{ˉ}$ . The correct answer is $cos θ - i sin θ$ .
Method #2Approach 2
Step 1: Identify what is being asked
We need the multiplicative inverse of $w = cos θ + i sin θ$ , which must satisfy $w \cdot \frac{1}{w} = 1$ .
Step 2: Eliminate $\cos\theta + i\sin\theta$
If $\frac{1}{w} = cos θ + i sin θ = w$ , then $w^{2} = 1$ , meaning $θ = 0$ or $θ = π$ . But $0 < θ < π$ , so this only works at boundary values — eliminate.
Step 3: Eliminate $-\cos\theta + i\sin\theta$
Multiplying $w \cdot (- cos θ + i sin θ) = - cos^{2} θ + i sin θ cos θ + i sin θ cos θ - sin^{2} θ = - (cos^{2} θ + sin^{2} θ) + 2 i sin θ cos θ = - 1 + i sin 2 θ \neq = 1$ . Eliminate.
Step 4: Eliminate $-\cos\theta - i\sin\theta$
This equals $- w$ , so $w \cdot (- w) = - w^{2} = - (cos 2 θ + i sin 2 θ) \neq = 1$ in general. Eliminate.
Step 5: Select $\cos\theta - i\sin\theta$
Checking: $(cos θ + i sin θ) (cos θ - i sin θ) = cos^{2} θ + sin^{2} θ = 1$ ✓. This is the correct answer.
Question 2
Let $w = cos θ + i sin θ$ where $0 < θ < π$ . What is $ar g (w^{2})$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $2 θ$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recall the multiplication rule for arguments
For any two complex numbers, $ar g (z_{1} z_{2}) = ar g (z_{1}) + ar g (z_{2})$ . Applying this to $w^{2} = w \cdot w$ :
Step 2: Apply the argument addition rule
$ar g (w^{2}) = ar g (w) + ar g (w) = 2 ar g (w) = 2 θ$
Step 3: Verify using Euler form
Since $w = e^{i θ}$ , we have $w^{2} = e^{2 i θ} = cos 2 θ + i sin 2 θ$ , so $ar g (w^{2}) = 2 θ$ .
Step 4: Confirm
The argument of $w^{2}$ is $2 θ$ , consistent with the general rule that squaring doubles the argument.
Method #2Approach 2
Step 1: What is being asked
We need the argument of $w^{2}$ given $w = e^{i θ}$ with $0 < θ < π$ .
Step 2: Eliminate $\theta^2$
The argument is an angle, not a product of $θ$ with itself. Arguments combine by addition under multiplication, not by multiplying the arguments together. Eliminate $θ^{2}$ .
Step 3: Eliminate $\dfrac{\theta}{2}$
$\frac{θ}{2}$ would be the argument of $w^{1/2}$ (a square root of $w$ ), not of $w^{2}$ . Squaring multiplies the exponent, not halves it. Eliminate.
Step 4: Eliminate $\theta + \pi$
Adding $π$ to the argument corresponds to multiplying by $e^{iπ} = - 1$ , i.e., negating $w$ . This is not the same as squaring $w$ . Eliminate.
Step 5: Select $2\theta$
Squaring doubles the argument: $ar g (w^{2}) = 2 θ$ . This follows directly from $e^{i θ} \cdot e^{i θ} = e^{2 i θ}$ .

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Step-by-step walkthrough

Step 1: Recognise the form of $w$

Step 2: Compute the reciprocal

Step 3: Use even/odd identities

Step 4: Confirm the answer

Step 1: Identify what is being asked

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

Step 3: Eliminate $-\cos\theta + i\sin\theta$

Step 4: Eliminate $-\cos\theta - i\sin\theta$

Step 5: Select $\cos\theta - i\sin\theta$

Step-by-step walkthrough

Step 1: Recall the multiplication rule for arguments

Step 2: Apply the argument addition rule

Step 3: Verify using Euler form

Step 4: Confirm

Step 1: What is being asked

Step 2: Eliminate $\theta^2$

Step 3: Eliminate $\dfrac{\theta}{2}$

Step 4: Eliminate $\theta + \pi$

Step 5: Select $2\theta$

13 more questions in this topic

Step-by-step walkthrough

Step 1: Recognise the form of $w$

Step 2: Compute the reciprocal

Step 3: Use even/odd identities

Step 4: Confirm the answer

Step 1: Identify what is being asked

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

Step 3: Eliminate $-\cos\theta + i\sin\theta$

Step 4: Eliminate $-\cos\theta - i\sin\theta$

Step 5: Select $\cos\theta - i\sin\theta$

Step-by-step walkthrough

Step 1: Recall the multiplication rule for arguments

Step 2: Apply the argument addition rule

Step 3: Verify using Euler form

Step 4: Confirm

Step 1: What is being asked

Step 2: Eliminate $\theta^2$

Step 3: Eliminate $\dfrac{\theta}{2}$

Step 4: Eliminate $\theta + \pi$

Step 5: Select $2\theta$

13 more questions in this topic