Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To solve these problems, we use De Moivre's Theorem, which states that for a complex number expressed in polar form [tex]\( z = r(\cos \theta + i \sin \theta) \)[/tex] and for a positive integer [tex]\( n \)[/tex], we have:
[tex]\[ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \][/tex]
Let's start with each complex number one by one:
1. [tex]\( (1+i)^5 \)[/tex]
- Convert [tex]\(1+i\)[/tex] into polar form. The magnitude is [tex]\( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \frac{\pi}{4} \)[/tex] since [tex]\( \tan^{-1}(1/1) = \frac{\pi}{4} \)[/tex].
- Using De Moivre’s Theorem: [tex]\( (\sqrt{2}(\cos{\frac{\pi}{4}} + i \sin{\frac{\pi}{4}}))^5 = (\sqrt{2})^5 (\cos{\frac{5\pi}{4}} + i \sin{\frac{5\pi}{4}}) \)[/tex].
- This simplifies to [tex]\( 4(\cos{\frac{5\pi}{4}} + i \sin{\frac{5\pi}{4}}) = 4(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i) = -4 - 4i \)[/tex].
So, [tex]\( (1+i)^5 \)[/tex] simplifies to [tex]\( -4-4i \)[/tex].
2. [tex]\( (-1+i)^6 \)[/tex]
- Convert [tex]\((-1+i)\)[/tex] into polar form. The magnitude is [tex]\( r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \frac{3\pi}{4} \)[/tex] since [tex]\( \tan^{-1}(-1/1) = \frac{3\pi}{4} \)[/tex].
- Using De Moivre’s Theorem: [tex]\( (\sqrt{2}(\cos{\frac{3\pi}{4}} + i \sin{\frac{3\pi}{4}}))^6 = (\sqrt{2})^6 (\cos{\frac{18\pi}{4}} + i \sin{\frac{18\pi}{4}}) \)[/tex].
- [tex]\(\frac{18\pi}{4} = 4\pi + \frac{\pi}{2}\)[/tex]. Using periodicity of [tex]\( \cos \)[/tex] and [tex]\( \sin \)[/tex], this simplifies to [tex]\( 4\pi + \frac{\pi}{2} = \frac{\pi}{2} \)[/tex].
- This simplifies to [tex]\( 8i \)[/tex].
So, [tex]\( (-1+i)^6 \)[/tex] simplifies to [tex]\( 8i \)[/tex].
3. [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex]
- Using De Moivre’s Theorem directly: [tex]\( 2((\cos{20^\circ} + i \sin{20^\circ})^3) = 2(\cos{60^\circ} + i \sin{60^\circ}) \)[/tex].
- [tex]\( \cos 60^\circ = \frac{1}{2} \)[/tex] and [tex]\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)[/tex].
- This simplifies to [tex]\( 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 + i\sqrt{3} \)[/tex].
So, [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex] simplifies to [tex]\( 1 + i\sqrt{3} \)[/tex].
4. [tex]\( 2[\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]^4 \)[/tex]
- Using De Moivre’s Theorem directly: [tex]\( 2((\cos{\frac{\pi}{4}} + i \sin{\frac{\pi}{4}})^4) = 2(\cos{\pi} + i \sin{\pi}) \)[/tex].
- [tex]\(\cos{\pi} = -1\)[/tex] and [tex]\(\sin{\pi} = 0\)[/tex].
- This simplifies to [tex]\( 2(-1 + i \cdot 0) = -2 \)[/tex].
Combining with other results, each complex number matches the simplified form as described.
So the correct pairs are:
- [tex]\( (1+i)^5 \)[/tex] simplifies to [tex]\( -4-4i \)[/tex].
- [tex]\( (-1+i)^6 \)[/tex] simplifies to [tex]\( 8i \)[/tex].
- [tex]\( 2[\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]^4 \)[/tex] simplifies to [tex]\( -2 \)[/tex].
- [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex] simplifies to [tex]\( 1 + i \sqrt{3} \)[/tex].
Note that [tex]\( 2(\sqrt{3}+i)^{10} \)[/tex] does not have a matching simplified form from the given choices, so it remains unmatched.
[tex]\[ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \][/tex]
Let's start with each complex number one by one:
1. [tex]\( (1+i)^5 \)[/tex]
- Convert [tex]\(1+i\)[/tex] into polar form. The magnitude is [tex]\( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \frac{\pi}{4} \)[/tex] since [tex]\( \tan^{-1}(1/1) = \frac{\pi}{4} \)[/tex].
- Using De Moivre’s Theorem: [tex]\( (\sqrt{2}(\cos{\frac{\pi}{4}} + i \sin{\frac{\pi}{4}}))^5 = (\sqrt{2})^5 (\cos{\frac{5\pi}{4}} + i \sin{\frac{5\pi}{4}}) \)[/tex].
- This simplifies to [tex]\( 4(\cos{\frac{5\pi}{4}} + i \sin{\frac{5\pi}{4}}) = 4(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i) = -4 - 4i \)[/tex].
So, [tex]\( (1+i)^5 \)[/tex] simplifies to [tex]\( -4-4i \)[/tex].
2. [tex]\( (-1+i)^6 \)[/tex]
- Convert [tex]\((-1+i)\)[/tex] into polar form. The magnitude is [tex]\( r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)[/tex].
- The argument [tex]\( \theta \)[/tex] is [tex]\( \frac{3\pi}{4} \)[/tex] since [tex]\( \tan^{-1}(-1/1) = \frac{3\pi}{4} \)[/tex].
- Using De Moivre’s Theorem: [tex]\( (\sqrt{2}(\cos{\frac{3\pi}{4}} + i \sin{\frac{3\pi}{4}}))^6 = (\sqrt{2})^6 (\cos{\frac{18\pi}{4}} + i \sin{\frac{18\pi}{4}}) \)[/tex].
- [tex]\(\frac{18\pi}{4} = 4\pi + \frac{\pi}{2}\)[/tex]. Using periodicity of [tex]\( \cos \)[/tex] and [tex]\( \sin \)[/tex], this simplifies to [tex]\( 4\pi + \frac{\pi}{2} = \frac{\pi}{2} \)[/tex].
- This simplifies to [tex]\( 8i \)[/tex].
So, [tex]\( (-1+i)^6 \)[/tex] simplifies to [tex]\( 8i \)[/tex].
3. [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex]
- Using De Moivre’s Theorem directly: [tex]\( 2((\cos{20^\circ} + i \sin{20^\circ})^3) = 2(\cos{60^\circ} + i \sin{60^\circ}) \)[/tex].
- [tex]\( \cos 60^\circ = \frac{1}{2} \)[/tex] and [tex]\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)[/tex].
- This simplifies to [tex]\( 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 + i\sqrt{3} \)[/tex].
So, [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex] simplifies to [tex]\( 1 + i\sqrt{3} \)[/tex].
4. [tex]\( 2[\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]^4 \)[/tex]
- Using De Moivre’s Theorem directly: [tex]\( 2((\cos{\frac{\pi}{4}} + i \sin{\frac{\pi}{4}})^4) = 2(\cos{\pi} + i \sin{\pi}) \)[/tex].
- [tex]\(\cos{\pi} = -1\)[/tex] and [tex]\(\sin{\pi} = 0\)[/tex].
- This simplifies to [tex]\( 2(-1 + i \cdot 0) = -2 \)[/tex].
Combining with other results, each complex number matches the simplified form as described.
So the correct pairs are:
- [tex]\( (1+i)^5 \)[/tex] simplifies to [tex]\( -4-4i \)[/tex].
- [tex]\( (-1+i)^6 \)[/tex] simplifies to [tex]\( 8i \)[/tex].
- [tex]\( 2[\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]^4 \)[/tex] simplifies to [tex]\( -2 \)[/tex].
- [tex]\( 2[\cos(20^\circ) + i \sin(20^\circ)]^3 \)[/tex] simplifies to [tex]\( 1 + i \sqrt{3} \)[/tex].
Note that [tex]\( 2(\sqrt{3}+i)^{10} \)[/tex] does not have a matching simplified form from the given choices, so it remains unmatched.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
