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then a and b are respectively :- a)64 and - 64√3
- b)128 and 128√3
- c)512 and - 512√3
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
then a and b are respectively :a)64 and - 64√3b)128 and 128&radi...
(√3 + i)10 = a + ib
Z = √3 + i = rcosθ + i rsinθ
⇒ √3 = rcosθ i = rsinθ
⇒ (√3)2 + (1)2 = r2cos2θ + r2sin2θ
⇒ 4 = r2
⇒ r = 2
tan = 1/√3
⇒ tan π/6
Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
(Z)10 = √3 + i = (2cos π/6 + 2i sin π/6)10
= 210 (cos π/6 + i sin π/6)10
210 (cos 10π/6 + i sin 10π/6)
= 210 (cos(2π - π/3) + i sin(2π - π/3))
= 210 (cos π/3 - i sin π/3)
= 210 (1/2 - i√3/2)
29(1 - i√3)
a = 29 = 512
b = - 29(√3) = -512√3
Z = √3 + i = rcosθ + i rsinθ
⇒ √3 = rcosθ i = rsinθ
⇒ (√3)2 + (1)2 = r2cos2θ + r2sin2θ
⇒ 4 = r2
⇒ r = 2
tan = 1/√3
⇒ tan π/6
Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
(Z)10 = √3 + i = (2cos π/6 + 2i sin π/6)10
= 210 (cos π/6 + i sin π/6)10
210 (cos 10π/6 + i sin 10π/6)
= 210 (cos(2π - π/3) + i sin(2π - π/3))
= 210 (cos π/3 - i sin π/3)
= 210 (1/2 - i√3/2)
29(1 - i√3)
a = 29 = 512
b = - 29(√3) = -512√3
Most Upvoted Answer
then a and b are respectively :a)64 and - 64√3b)128 and 128&radi...
Understanding the Problem
We need to compute \((\sqrt{3} + i)^{10}\) and express it in the form \(a + ib\) where \(a, b \in \mathbb{R}\).
Using Polar Form
1. **Convert to Polar Coordinates**:
- The modulus \(r\) is given by:
\[
r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2
\]
- The argument \(\theta\) is:
\[
\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}
\]
2. **Express in Polar Form**:
\[
\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)
\]
Using De Moivre's Theorem
3. **Raise to the Power of 10**:
\[
(\sqrt{3} + i)^{10} = 2^{10}\left(\cos\left(10 \cdot \frac{\pi}{6}\right) + i\sin\left(10 \cdot \frac{\pi}{6}\right)\right)
\]
4. **Calculate \(2^{10}\)**:
\[
2^{10} = 1024
\]
5. **Calculate the Angles**:
- The angle simplifies to:
\[
10 \cdot \frac{\pi}{6} = \frac{5\pi}{3}
\]
- Now, using the unit circle:
\[
\cos\frac{5\pi}{3} = \frac{1}{2}, \quad \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}
\]
Final Calculation
6. **Combine Results**:
\[
(\sqrt{3} + i)^{10} = 1024\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 512 - 512i\sqrt{3}
\]
Thus, \(a = 512\) and \(b = -512\sqrt{3}\). The correct answer is:
Answer
- \(a = 512\)
- \(b = -512\sqrt{3}\)
Hence, the option is **C**.
We need to compute \((\sqrt{3} + i)^{10}\) and express it in the form \(a + ib\) where \(a, b \in \mathbb{R}\).
Using Polar Form
1. **Convert to Polar Coordinates**:
- The modulus \(r\) is given by:
\[
r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2
\]
- The argument \(\theta\) is:
\[
\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}
\]
2. **Express in Polar Form**:
\[
\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)
\]
Using De Moivre's Theorem
3. **Raise to the Power of 10**:
\[
(\sqrt{3} + i)^{10} = 2^{10}\left(\cos\left(10 \cdot \frac{\pi}{6}\right) + i\sin\left(10 \cdot \frac{\pi}{6}\right)\right)
\]
4. **Calculate \(2^{10}\)**:
\[
2^{10} = 1024
\]
5. **Calculate the Angles**:
- The angle simplifies to:
\[
10 \cdot \frac{\pi}{6} = \frac{5\pi}{3}
\]
- Now, using the unit circle:
\[
\cos\frac{5\pi}{3} = \frac{1}{2}, \quad \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}
\]
Final Calculation
6. **Combine Results**:
\[
(\sqrt{3} + i)^{10} = 1024\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 512 - 512i\sqrt{3}
\]
Thus, \(a = 512\) and \(b = -512\sqrt{3}\). The correct answer is:
Answer
- \(a = 512\)
- \(b = -512\sqrt{3}\)
Hence, the option is **C**.
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then a and b are respectively :a)64 and - 64√3b)128 and 128√3c)512 and - 512√3d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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