-14+102i

\(N(a+bi)=(a+bi)(a-bi)=a^{2}+b^{2}\)

N(-14+102i)=(-14+102i)(-14-102i)

\(=(-14)^{2}+(102)^{2}\)

=10600

Then, the prime factorization becomes

\(10600 = 2\times 2\times 2\times 5\times 5\times 53\)

\(=[(1+i)(1-i)]^{3}[(1+2i)(1-2i)]^{2}(7+2i)(7-2i)\)

\(N(a+bi)=(a+bi)(a-bi)=a^{2}+b^{2}\)

N(-14+102i)=(-14+102i)(-14-102i)

\(=(-14)^{2}+(102)^{2}\)

=10600

Then, the prime factorization becomes

\(10600 = 2\times 2\times 2\times 5\times 5\times 53\)

\(=[(1+i)(1-i)]^{3}[(1+2i)(1-2i)]^{2}(7+2i)(7-2i)\)