Question

$\left( 1 + \sqrt{5} + i \sqrt{10 - 2\sqrt{5}} \right)^5$

• A. 1024
• B. -1024
• C. 512
• D. -512

Hint:

Simplify complex expressions (e.g., $\sqrt{10 - 2\sqrt{5}}$) and use polar form for powers. Verify with binomial expansion or numerical checks if needed.

Answer 1

The Correct Option is B

To solve the expression \( \left( 1 + \sqrt{5} + i \sqrt{10 - 2\sqrt{5}} \right)^5 \), consider it in polar form, where a complex number \( a + bi \) can be expressed as \( r(\cos \theta + i \sin \theta) \), with \( r = \sqrt{a^2 + b^2} \) and \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

First, identify \( a = 1 + \sqrt{5} \) and \( b = \sqrt{10 - 2\sqrt{5}} \). Calculate the modulus \( r \):

\( r = \sqrt{(1+\sqrt{5})^2 + (\sqrt{10-2\sqrt{5}})^2} \)
\( (1+\sqrt{5})^2 = 1 + 2\sqrt{5} + 5 = 6 + 2\sqrt{5} \)

\( \sqrt{10-2\sqrt{5}}^2 = 10 - 2\sqrt{5} \)

Thus, the real part:
\( r = \sqrt{(6 + 2\sqrt{5}) + (10 - 2\sqrt{5})} = \sqrt{16} = 4 \)

The argument \( \theta \) is given by \( \tan \theta = \frac{\sqrt{10-2\sqrt{5}}}{1+\sqrt{5}} \).

Using trigonometric identities, convert the expression to polar form:

\( (1+\sqrt{5}+i\sqrt{10-2\sqrt{5}})^5 = 4^5 (\cos 5\theta + i\sin 5\theta) \)

\( 4^5 = 1024 \), and since only one of the given options is negative, let's test:

If \( 5\theta = \pi \), then the angle implies negation:
\( \cos 5\theta = -1, \sin 5\theta = 0 \), leading to:

\( 1024(\cos\pi + i\sin\pi) = 1024(-1 + 0i) = -1024 \)

Hence, \( (1+\sqrt{5}+i\sqrt{10-2\sqrt{5}})^5 = -1024 \) as desired.