$$(1+i)^{102} \quad?$$
$$(1+i)^{102} = \left((1+i)^2\right)^{51}$$
$$\\(1+i)^2=1^2+2*1*i+i^2=1+2i+i^2 \quad | \quad \textcolor[rgb]{0,0,1}{i^2 = \left(\sqrt{-1}\right)^2 = -1} \\
(1+i)^2=1+2i-1\\
\textcolor[rgb]{1,0,0}{(1+i)^2=2i
}$$
$$(1+i)^{102}=(2i)^{51}
=2^{51}\times i^{51}$$
$$\\i^{51} = \left(i^4\right)^{12}\times (i^3)$$
$$\\i^4=i^2*i^2=(-1)*(-1)=1\\
i^3=i^2*i=(-1)*i=-i$$
$$\\i^{51} = \left(1\right)^{12}\times (-i) =-i$$
$$(1+i)^{102} =2^{51}\times (-i) =-2^{51}i$$
$$\boxed{(1+i)^{102}=-2^{51}i=-2251799813685248i}$$
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