Question

Find the value:

$(1+i{)}^{12}+(1-i{)}^{12}$, $i=\sqrt{-1}$.

Answer 1

$(1+i{)}^{12}+(1-i{)}^{12}$

Let's solve this problem using the division algorithm as it is easier compared to other methods

The algorithm is:

$Divident=Divisor\times Quotient+Remainder$

We can directly write the solution for $(1+i{)}^n\pm (1-i{)}^n,$ as

$(-4{)}^q\times (1+i{)}^r\pm (-4{)}^q\times (1-i{)}^r$

where, $q$ and $r$ are the quotient and remainder, respectively, when we divide $n$ by $4$

[NOTE : Dividing the power $n$ by $4$ is part of this method and hence, we will always divide the power by $4$]

Here, $n=12$

$\therefore usingdivisionalgorithm,12=4\times 3+0$

$\therefore q=3andr=0$

$\therefore (-4{)}^3\times (1+i{)}^0+(-4{)}^3\times (1-i{)}^0$

$\Rightarrow (-64)+(-64)$

$\Rightarrow -128$

$\therefore$ The answer is $-128$