Problem 35
Question
$$\left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)^{12}\left[2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^{5}=2^{5}\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)$$ $$\begin{array}{l} =32\left[\cos \left(\frac{4 \pi}{3}+\frac{5 \pi}{6}\right)+i \sin \left(\frac{4 \pi}{3}+\frac{5 \pi}{6}\right)\right] \\ =32\left(\cos \frac{13 \pi}{6}+i \sin \frac{13 \pi}{6}\right)=32\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)=16 \sqrt{3}+16 i \end{array}$$
Step-by-Step Solution
Verified Answer
The expression equals \( 16\sqrt{3} + 16i \).
1Step 1: Identify given expressions
We begin with the equation expressed in trigonometric form with complex numbers: \( \left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)^{12}\left[2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^{5} \). Note the expression includes De Moivre's Theorem.
2Step 2: Apply De Moivre's Theorem
Use De Moivre's Theorem to simplify \( \left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)^{12} \). According to the theorem, \( (\cos \theta + i \sin \theta)^n = \cos (n\theta) + i\sin (n\theta) \). Substitute \( \theta = \frac{\pi}{9} \) and \( n = 12 \), yielding \( \cos (12 \times \frac{\pi}{9}) + i\sin (12 \times \frac{\pi}{9}) = \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} \).
3Step 3: Simplify the second expression
Apply De Moivre's Theorem again for the expression \( \left[2(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6})\right]^5 \). Break it as \( 2^5 \times (\cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6}) \) by using \( \theta = \frac{\pi}{6} \) and \( n = 5 \). This results in \( 32(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) \).
4Step 4: Combine the expressions
Multiply the simplified expressions from Step 2 and Step 3: \( (\cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3}) \times 32(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) \).
5Step 5: Add the angles
Combine the angles: \( \cos \left(\frac{4\pi}{3} + \frac{5\pi}{6}\right) + i \sin \left(\frac{4\pi}{3} + \frac{5\pi}{6}\right) = \cos \frac{13\pi}{6} + i \sin \frac{13\pi}{6} \). Simplifying \( \frac{13\pi}{6} \) into the principal value gives \( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \).
6Step 6: Evaluate the final expression
Finally, multiply \( 32 \) by the evaluated trigonometric form, \( 32(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) \). This results in \( 32 \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \), simplifying to \( 16\sqrt{3} + 16i \).
Key Concepts
De Moivre's TheoremTrigonometric FormPrincipal ValueComplex Number Multiplication
De Moivre's Theorem
De Moivre's Theorem is a cornerstone of complex number arithmetic, especially in simplifying expressions involving powers of complex numbers. It establishes a connection between complex numbers in trigonometric form and their corresponding powers. The theorem states: for a complex number in trigonometric form \((\cos \theta + i \sin \theta)\), raising it to the power \(n\), results in \(\cos(n \theta) + i \sin(n \theta)\). This method simplifies calculations significantly when dealing with higher powers.
For example, in the given problem, we have \(\left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)^{12}\). By applying De Moivre’s Theorem, we substitute \(\theta = \frac{\pi}{9}\) and \(n = 12\), transforming it into \(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\). This reduction simplifies the expression immensely and highlights the power of the theorem in dealing with complex numbers.
Understanding this theorem is crucial because it not only simplifies calculations but also provides a clear way to visualize the multiplication and power of complex numbers.
For example, in the given problem, we have \(\left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)^{12}\). By applying De Moivre’s Theorem, we substitute \(\theta = \frac{\pi}{9}\) and \(n = 12\), transforming it into \(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\). This reduction simplifies the expression immensely and highlights the power of the theorem in dealing with complex numbers.
Understanding this theorem is crucial because it not only simplifies calculations but also provides a clear way to visualize the multiplication and power of complex numbers.
Trigonometric Form
The trigonometric form represents complex numbers in a way that is particularly useful for multiplication and division. Instead of using the standard form \(a + bi\), we utilize the expression \(r(\cos \theta + i \sin \theta)\), where \(r\) is the modulus and \(\theta\) is the argument of the complex number.
This can be particularly powerful when performing operations such as multiplication or division. When multiplying complex numbers in trigonometric form, you simply multiply their moduli and add their angles. For division, divide the moduli and subtract the arguments.
In the example exercise, the complex numbers are expressed as \((\cos \theta + i \sin \theta)\), ready to apply De Moivre's Theorem. This expression greatly simplifies the problem when raised to a power and is crucial for understanding the principles behind complex number operations.
This can be particularly powerful when performing operations such as multiplication or division. When multiplying complex numbers in trigonometric form, you simply multiply their moduli and add their angles. For division, divide the moduli and subtract the arguments.
In the example exercise, the complex numbers are expressed as \((\cos \theta + i \sin \theta)\), ready to apply De Moivre's Theorem. This expression greatly simplifies the problem when raised to a power and is crucial for understanding the principles behind complex number operations.
Principal Value
Understanding the principal value of a complex number is key to working with trigonometric forms and De Moivre's Theorem. The principal value refers to the smallest positive angle in which a periodic function like sine or cosine returns to its initial value. It ensures a unique representation of the trigonometric form.
In terms of complex numbers, finding the principal value means adjusting an angle so that it lies within a standard range, usually between \(-\pi\) and \(\pi\) or \(0\) and \(2\pi\).
In our exercise, after applying De Moivre’s Theorem and simplifying the powers, the resulting angle \(\frac{13\pi}{6}\) was outside the principal range. By reducing \(\frac{13\pi}{6}\) appropriately, we obtained \(\frac{\pi}{6}\), ensuring that our final results are within the expected principal value range, making the calculations consistent and correct.
In terms of complex numbers, finding the principal value means adjusting an angle so that it lies within a standard range, usually between \(-\pi\) and \(\pi\) or \(0\) and \(2\pi\).
In our exercise, after applying De Moivre’s Theorem and simplifying the powers, the resulting angle \(\frac{13\pi}{6}\) was outside the principal range. By reducing \(\frac{13\pi}{6}\) appropriately, we obtained \(\frac{\pi}{6}\), ensuring that our final results are within the expected principal value range, making the calculations consistent and correct.
Complex Number Multiplication
Multiplication of complex numbers becomes particularly intuitive when using the trigonometric form. This involves two main steps: multiplying the magnitudes (moduli) and adding the angles (arguments).
In our example, after finding that \(\frac{13\pi}{6}\) simplifies to \(\frac{\pi}{6}\), we multiply the initial factor \(32\) by \(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\). This gives us \(32\left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right)\). By performing the multiplication, we arrive at the expression \(16\sqrt{3} + 16i\).
This step emphasizes how simpler multiplication of moduli and addition of angles yield direct results that might be cumbersome using algebraic manipulation. Understanding complex number multiplication in trigonometric form provides a structured way to handle complex operations efficiently.
In our example, after finding that \(\frac{13\pi}{6}\) simplifies to \(\frac{\pi}{6}\), we multiply the initial factor \(32\) by \(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\). This gives us \(32\left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right)\). By performing the multiplication, we arrive at the expression \(16\sqrt{3} + 16i\).
This step emphasizes how simpler multiplication of moduli and addition of angles yield direct results that might be cumbersome using algebraic manipulation. Understanding complex number multiplication in trigonometric form provides a structured way to handle complex operations efficiently.
Other exercises in this chapter
Problem 33
\(2 x+2 y i=-9+2 i\) implies \(2 x=-9\) and \(2 y=2 .\) Hence \(z=-\frac{9}{2}+i\)
View solution Problem 34
\(\left(z^{4}-1\right)^{2}=0\) is the same as \((z-i)^{2}(z+i)^{2}(z-1)^{2}(z+1)^{2}=0 .\) Thus \(z_{1}=1, z_{2}=-1, z_{3}=i,\) and \(z_{4}=-i\) are roots of mu
View solution Problem 36
$$\frac{1}{z}=\ln (-1)=\log _{e} 1+i(\pi+2 n \pi)=(2 n+1) \pi i \text { and so } z=-\frac{i}{(2 n+1) \pi}.$$
View solution Problem 36
$$\frac{\left[8\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)\right]^{3}}{\left[2\left(\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}\right)\right]^{10}}=\
View solution