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\left(\sqrt{2}+i\right)^{2}
Multiply \sqrt{2}+i and \sqrt{2}+i to get \left(\sqrt{2}+i\right)^{2}.
\left(\sqrt{2}\right)^{2}+2i\sqrt{2}-1
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+i\right)^{2}.
2+2i\sqrt{2}-1
The square of \sqrt{2} is 2.
1+2i\sqrt{2}
Subtract 1 from 2 to get 1.