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If $\phi = \sqrt {n + \sqrt {n + \sqrt {n + ...}}}$ , where $\phi$ represents $\frac {1 + \sqrt {5}}{2}$ , and $\frac {n + ni}{3n - ni}$ is a complex number which can be solved to equal $\frac {a + bi}{c}$ , where $i^2$ equals $-1$ and $a$ , $b$ and $c$ are positive integers, find the value of $(a + b + c)^{\frac {c}{a + b}}$ ?

The answer is 32.

1 solution

Tom Engelsman
Jun 5, 2021

If we solve the nested radical for $n$ , we obtain:

$\phi = \sqrt{n+\sqrt{n+\sqrt{n+...}}} \Rightarrow \phi^{2} - n = \sqrt{n+\sqrt{n+\sqrt{n+...}}} \Rightarrow \phi^{2} - \phi = n \Rightarrow \frac{3+\sqrt{5}}{2} - \frac{1+\sqrt{5}}{2} = n \Rightarrow n = 1.$

If $\large z = \frac{n+ni}{3n-ni}$ , then $\large z = \frac{1+i}{3-i} \cdot \frac{3+i}{3+i} = \frac{3-1+i+3i}{9-i^2} = \frac{2+4i}{10} = \frac{1+2i}{5}$ . Thus $a=1, b =2, c=5$ , and we end up with:

$\large (a+b+c)^{c/(a+b)} = (1+2+5)^{5/(1+2)} = (8^{1/3})^5 = 2^5 = \boxed{32}.$

Phiiii-dom

The answer is 32.

1 solution

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