Answer :
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.
