Find the value of $$\left(\frac{-1+\sqrt 3i}{2}\right)^{15} + \left(\frac{-1-\sqrt 3i}{2}\right)^{15}.$$
In general, how do we find the value of expansion of equation of high orders other than binomial expansion?
[Math] Expansion of complex equation.
binomial theoremcomplex numbers
Related Solutions
For a general $n$, the constant term in $$f(x) = (1 + x/5)^n (2-3/x)^2$$ can be found by observing first that $$(2-3/x)^2 = 2 - 12x^{-1} + 9x^{-2}.$$ Then we see that the constant term of $f$ is given by $$1 \cdot 2 + \binom{n}{1} (x/5)(-12x^{-1}) + \binom{n}{2}(x/5)^2(9x^{-2}),$$ since these are the only terms for which the power of $x$ will be zero.
Here we look at some arguments why taking out a factor $2.5$ is a convenient choice.
Problem: We want to (manually) approximate $\sqrt{6}$ by using the first few terms of the binomial series expansion of \begin{align*} \sqrt{1-4x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{4}\\ &= 1-2x-2x^2-4x^3+\cdots\tag{1} \end{align*}
In order to apply (1) we are looking for a number $y$ with \begin{align*} \sqrt{1-4x}&=\sqrt{6y^2}=y\sqrt{6}\tag{2}\\ \color{blue}{\sqrt{6}}&\color{blue}{=\frac{1}{y}\sqrt{1-4x}} \end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root. We obtain from (2) \begin{align*} 1-4x&=6y^2\\ 4x&=1-6y^2\\ \color{blue}{x}&\color{blue}{=\frac{1}{4}-\frac{3}{2}y^2}\tag{3} \end{align*}
When looking for a nice $y$ which fulfills (3) there are some aspects to consider:
We have to respect the radius of convergence $|x|<\frac{1}{4}$.
Since we want to calculate an approximation of $\sqrt{6}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.
Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{6}$.
We will see it's not hard to find values which have these properties.
We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that
\begin{align*}
\frac{3}{2}y^2
\end{align*}
is close to $\frac{1}{4}$. So, $y^2$ is close to $\frac{1}{4}\cdot\frac{2}{3}=\frac{1}{6}$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $a,b$ so that
\begin{align*}
y^2=\frac{a^2}{b^2}\approx \frac{1}{6}
\end{align*}
This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $6$ we might quickly come to $100$ and $16$. These are two small squares and we have $6\cdot 16=96$ close to $100$. That's all.
Now it's time to harvest. We choose $y^2=\frac{16}{100}=\frac{4}{25}$ resp. $y=\frac{2}{5}$. We obtain for $x$ from (3) \begin{align*} x=\frac{1}{4}-\frac{3}{2}y^2=\frac{1}{4}-\frac{3}{2}\cdot\frac{4}{25}=\frac{1}{100} \end{align*} We have now an appropriate value $x=\frac{1}{100}$ and we finally get from (1) the approximation: \begin{align*} \color{blue}{\sqrt{6}} \approx \frac{5}{2}\left(1-2\cdot 10^{-2}-2\cdot 10^{-4}- 4\cdot 10^{-4}\right)\color{blue}{=2.449\,4}9 \end{align*}
We have $\sqrt{6}=\color{blue}{2.449\,4}89\,7\ldots$ with an approximation error $\approx 2.572\times 10^{-7}$. This result is quite impressive when considering that we have used just four terms of the binomial series.
At this point it might be clear that factoring out $2.5=\frac{5}{2}=\frac{1}{y}$ is a good choice.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
Best Answer
In general, if you want to find powers of a complex number, write it in polar form i.e. in the form of $r e^{i \theta} $ so that $(r e^{i \theta})^n = r^n e^{i n \theta} $. Then you can convert it back to $a + ib$ form easily as $r^n \cos(n \theta) + i r^n \sin(n \theta) $.