Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
 A: We need to solve for $ab=2$ and $a+b=1$
$\implies a,b=?$
Use $x^2+y^2=(x+y)^2-2xy$ twice to reach at $$a^4+b^4=1$$
A: Hints: 


*

*Just perform the operations. 

*Taking $4$th power is taking square twice. 

*It's enough to do it for one term and take its real part.

A: Alright, you want difference of squares have some difference of squares
$$(\frac{1+i \sqrt7}{2})^4 + (\frac{1-i \sqrt7}{2})^4=$$
Lets add $0$ cleverly
$$=(\frac{1+i \sqrt7}{2})^4 -2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2+ (\frac{1-i \sqrt7}{2})^4 + 2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2$$
$$=((\frac{1+i \sqrt7}{2})^2 - (\frac{1-i \sqrt7}{2})^2)^2 + 2(\frac{1+7}{4})^2 $$
The $2(\frac{1+7}{4})^2$ comes from the final term of the second line, it can be viewed as a difference of squares, squared. Next deal with the difference of tsquares in the squared parenthesis
$$ = (\frac{2\cdot2 i \sqrt{7}}{4})^2 + 8 $$
$$ =-7+8=1$$
QED
A: Let $(a_n)_n$ the sequence verifying $\begin{cases} a_0=2\\ a_1=1\\a_{n+2}=a_{n+1}-2a_n\end{cases}$
The characteristic equation of this linear recurrence relation is $x^2=x-2$ 
Whose roots are $\dfrac{1\pm i\sqrt{7}}2$.
Thus $a_n=\alpha r^n+\beta {\bar r}^n$ and given the initial conditions then $\alpha=\beta=1$.
So $a_n=\left(\dfrac{1+i\sqrt{7}}2\right)^n+\left(\dfrac{1-i\sqrt{7}}2\right)^n$
We are asked to calculate $a_4$ ?


*

*$a_2=a_1-2a_0=1-4=-3$

*$a_3=a_2-2a_1=-3-2=-5$

*$a_4=a_3-2a_2=-5+6=1$


Use the integer sequence to calculate any other power you desire...
A: Let $x=(1+i\sqrt 7\;)/(2\sqrt 2\;).$  Since $x\bar x=x^2\bar x^2=1$ and $x+\bar x=1/\sqrt 2\;$ we have $$((1+i\sqrt 7\;)/2)^4+(1-i\sqrt 7\;)/2)^4=4(x^4+ \bar x^4)=$$     $$=4((x^2+\bar x^2)^2-2x^2\bar x^2)=4((x^2+\bar x^2)^2-2)=$$    $$=4( ((x+\bar x)^2-2x\bar x)^2-2)=$$ $$=4((1/\sqrt 2\;)^2-2)^2-2)=4((1/2-2)^2-2)=4(9/4-2)=1.$$
A: Let $w = \dfrac{1 + i\sqrt{7}}{2}$, then $\dfrac{1 - i\sqrt{7}}{2} = \bar{w}$, its complex conjugate. We are required to find the value of $w^4 + \bar{w}^4$.  
Observer that
$$w \bar{w} = 2 \quad {\rm and} \quad w + \bar{w} = 1.$$
Now
$$(w + \bar{w})^2 = w^2 + 2w\bar{w} + \bar{w}^2 \quad \Rightarrow \quad w^2 + \bar{w}^2 = 1 - 2 w \bar{w} = 1 - 2(2) = -3.$$
Squaring the last expression we have
\begin{align*}
(w^2 + \bar{w}^2)^2 &= (-3)^2\\
w^4 + 2w^2 \bar{w}^2 + \bar{w}^4 &= 9\\
\Rightarrow w^4 + \bar{w}^4 &= 9 - 2(w \bar{w})^2\\
&= 9 - 2(2)^2\\
&= 9 - 8\\
&= 1,
\end{align*}
as required to show. 
A: Yet another way:   let $a=\frac{1+i\sqrt{7}}{2}$ and $b=\frac{1-i\sqrt{7}}{2}$ then $a+b=1, ab=2$ so $a,b$ are the roots of the quadratic $x^2-x+2=0\,$.
Therefore $a^2=a-2\,$, then successively:


*

*$a^3 = a\cdot a^2 = a \cdot(a-2) = a^2-2a = (a-2)-2a=-a-2$

*$a^4= a \cdot a^3 = a(-a-2) = -a^2-2a=-(a-2)-2a=-3a+2$
The same holds for $\,b\,$ so $\;a^4+b^4=(-3a+2)+(-3b+2)=-3(a+b)+4=-3\cdot 1+4=1\,$.
A: We have $\dfrac {1+i \sqrt 7}{2} =  \sqrt 2 \left(\cos \theta + i \sin \theta \right)$ where $ \cos \theta = \dfrac{1}{2 \sqrt 2}$
We need to calculate $2 \Re \left(\dfrac {1+i \sqrt 7}{2} \right)^4 = 2 \times (\sqrt 2)^4 \times \cos 4\theta = 8 \cos 4 \theta$
We have $\cos 2 \theta = 2 \cos^2 \theta -1 = -\dfrac{3}{4}$ and use that to get $\cos 4 \theta = \dfrac{1}{8}$
Hence we see that the sum evaluates to $8 \cos 4 \theta = 1$
