Sure, let's apply De Moivre's theorem to find the square of the given complex numbers step by step.
### Complex number (ii) [tex]\( 2 + 2\sqrt{3} i \)[/tex]:
1. Convert to polar form:
- Magnitude (r): [tex]\( r = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \)[/tex]
- Argument (θ): [tex]\( \theta = \tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)[/tex]
So, [tex]\( 2 + 2\sqrt{3}i = 4(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) \)[/tex].
2. Apply De Moivre's theorem:
[tex]\[
(4 \cdot (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}))^2 = 4^2 (\cos 2 \cdot \frac{\pi}{3} + i \sin 2 \cdot \frac{\pi}{3})
\][/tex]
[tex]\[
= 16 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})
\][/tex]
3. Evaluate:
[tex]\[
\cos \frac{2\pi}{3} = -\frac{1}{2}, \quad \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}
\][/tex]
[tex]\[
16 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 16 \cdot -\frac{1}{2} + 16 \cdot i \frac{\sqrt{3}}{2}
\][/tex]
[tex]\[
= -8 + 8\sqrt{3} i
\][/tex]
Thus, [tex]\( (2 + 2 \sqrt{3} i)^2 = -8 + 8\sqrt{3}i \)[/tex]. Numerically, we know it is approximately [tex]\( -8 + 13.85641i \)[/tex].
### Complex number (iii) [tex]\( -2 - 2 \sqrt{3} i \)[/tex]:
1. Convert to polar form:
- Magnitude (r): [tex]\( r = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = 4 \)[/tex]
- Argument (θ): [tex]\( \theta = \tan^{-1}\left(\frac{-2\sqrt{3}}{-2}\right) = \tan^{-1}(\sqrt{3}) \)[/tex]
Since it lies in the third quadrant, [tex]\(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \)[/tex]
So, [tex]\( -2 - 2\sqrt{3}i = 4(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) \)[/tex].
2. Apply De Moivre's theorem:
[tex]\[
(4 \cdot (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}))^2 = 16 (\cos \frac{8\pi}{3} + i \sin \frac{8\pi}{3})
\][/tex]
3. Evaluate:
[tex]\(\cos \frac{8\pi}{3} = -\frac{1}{2}, \quad \sin \frac{8\pi}{3} = \frac{\sqrt{3}}{2}\)[/tex]
[tex]\[
16 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -8 + 8\sqrt{3}i
\][/tex]
Thus, [tex]\( (-2 - 2 \sqrt{3} i)^2 = -8 + 8\sqrt{3}i \)[/tex]. Numerically, it is approximately [tex]\( -8 + 13.85641i \)[/tex].
### Complex number (v) [tex]\( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \)[/tex]:
1. Convert to polar form:
- Magnitude (r): [tex]\( r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1 \)[/tex]
- Argument (θ): [tex]\( \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(-\sqrt{3}) = \frac{2\pi}{3} \)[/tex]
So, [tex]\( -\frac{1}{2} + \frac{\sqrt{3}i}{2} = 1 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) \)[/tex].
2. Apply De Moivre's theorem:
[tex]\[
(1 \cdot (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}))^2 = 1 (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3})
\][/tex]
3. Evaluate:
[tex]\(\cos \frac{4\pi}{3} = -\frac{1}{2}, \quad \sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}\)[/tex]
[tex]\[
= -\frac{1}{2} - i \frac{\sqrt{3}}{2}
\][/tex]
Thus, [tex]\( \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \)[/tex]. Numerically, it is approximately [tex]\( -0.5 - 0.8660i \)[/tex].
To summarize:
- [tex]\((2 + 2 \sqrt{3} i)^2 = -8 + 13.85641i\)[/tex]
- [tex]\((-2 - 2 \sqrt{3} i)^2 = -8 + 13.85641i\)[/tex]
- [tex]\(\left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^2 = -0.5 - 0.8660i\)[/tex]
