Phiiii-dom

If we square the given expression we get:

$\phi^2=n+\sqrt{n+\sqrt{n+.....}}$ .Take $n$ to the left side and we have:

$\phi^2-n=\sqrt{n+\sqrt{n+.....}}$ .But $\sqrt{n+\sqrt{n+....}}=\phi$ , therefore our equation transforms into :

$\phi^2-n=\phi$ where $\phi^2=\phi+n$ .We know that $\phi$ satisfies the equation $\phi^2=\phi+1$ and equating this to our equation we easily get $\boxed{n=1}$ .From this we know our complex number is :

$\dfrac{1+i}{3-i}$ .Multiplying both the numerator and the denominator by $3+i$ our complex number bcomes :

$\dfrac{(1+i)(3+i)}{(3-i)(3+i)}=\dfrac{4i+2}{10}=\dfrac{2i+1}{5}$ .Thus $\boxed{a=1}$ , $\boxed{b=2}$ and $\boxed{c=5}$ .And we have $(a+b+c)^{\dfrac{c}{a+b}}=8^{\dfrac{5}{3}}=\boxed{32}$ .

What a nice final expression.

It is "quite known" that $\varphi = \sqrt{1+\sqrt{1+\sqrt{\cdots}}}$ . Thus $n =1$ .

Simplifying $\frac{1+i}{3-i}$ , we have $\frac{(1+i)(3+i)}{(3-i)(3+i)} = \frac{a+bi}{c} \Rightarrow \frac{2+4i}{10} = \frac{a+bi}{c}$ and thus $a = 1, b = 2, c= 5$ .

For it the final expression becomes $(1+2+5)^{\frac{5}{1+2}} \Rightarrow 2^{3 \cdot \frac{5}{3}} \Rightarrow \boxed{32.}$

Given that $\phi=\frac{1+\sqrt{5}}{2}$ and $\frac{n+ni}{3n-ni}$ is a complex no. which can be solved down to $\frac{a+bi}{c}$ . So, at first we have ---

$\phi = \sqrt{n+\sqrt{n+\sqrt{n+....}}}$ .....(i)

$\phi^2=n+ \sqrt{n+\sqrt{n+\sqrt{n+....}}}$ [Squaring both sides]

$\phi^2=n+ \phi$ [From (i)]

$(\frac{1+\sqrt{5}}{2})^2=n+(\frac{1+\sqrt{5}}{2})$

$\frac{1+2\sqrt{5}+5}{4}=\frac{2n+1+\sqrt{5}}{2}$

$\frac{2\sqrt{5}+6}{2}=2n+1+\sqrt{5} \implies 4n+2+2\sqrt{5}=6+2\sqrt{5} \implies 4n=4 \implies n=\boxed{4}$

Then, the complex no. in the question is $\frac{n+ni}{3n-ni}=\frac{1+i}{3-i}$ . Then, we solve this as follows --

$\frac{1+i}{3-i}$

$=\frac{(1+i)(3+i)}{(3-i)(3+i)}$

$=\frac{3+i+3i+i^2}{3^2-i^2}$ [Using the identity $a^2-b^2=(a+b)(a-b)$ ]

$=\frac{3+4i-1}{9+1}$ [since $i^2=(-1)$ ]

$=\frac{2+4i}{10} = \frac{2(1+2i)}{10} = \frac{1+2i}{5}$

So, we have the complex no. in $\frac{a+bi}{c}$ form with $a=1,b=2,c=5$ . Then, we have ---

$(a+b+c)^{\frac{c}{a+b}}=(1+2+5)^{\frac{5}{1+2}} = (8)^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = (2)^{3\times \frac{5}{3}} = 2^5= \boxed{32}$

$\phi = sqrt(n + sqrt(n + ...))$

Thus, $\phi = sqrt(n + \phi)$ .

Squaring both sides: $\phi^{2} = n + \phi$ .

Since $\phi^{2} = \phi + 1$ , we can conclude that $n = 1$ .

Our complex number is then $(1 + i)/(3 - i)$ . Multiplying this by $(3 + i)/(3 + i)$ yields: $(2 + 4i)/(10) = (1 + 2i)/5$ . You should have been more careful when you stated that the complex number can be represented as $(a + bi)/c$ , since there are infinite representations that are of the form $(na + nbi)/nc$ , $n$ a nonzero real number, which equal the complex number you gave us - and those will yield different results. I'll assume that you wanted us to consider a = 1, b = 2 and c = 5. Then the answer would be $(1 + 2 + 5)^{(5/(1 + 2))}$ , which equals $8^{5/3}$ , or $2^{5}$ , which is $32$