Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Simplify the complex numbers using de Moivre's Theorem and match them with their solutions.

- [tex](1+i)^5[/tex]
- [tex](-1+i)^6[/tex]
- [tex]2(\sqrt{3}+i)^{10}[/tex]
- [tex]2\left[\cos \left(20^{\circ}\right)+i \sin \left(20^{\circ}\right)\right]^3[/tex]
- [tex]2\left[\cos \left(\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}\right)\right]^4[/tex]

Solutions:
- [tex]8i[/tex]
- [tex]-4-4i[/tex]



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.