How do you multiply ${(3 + 2 i)}^{2}$ $(3+2i)^2$ ?

Question

15682 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-16T17:13:30

You remember that $i^{2} = - 1$ $i^2=-1$

${(3 + 2 i)}^{2} = (3 + 2 i) (3 + 2 i)$ $(3+2i)^2=(3+2i)(3+2i)$

$9 + 6 i + 6 i + 4 i^{2}$ $9+6i+6i+4i^2$

$9 + 12 i + 4 (- 1)$ $9+12i+4(-1)$

$9 + 12 i - 4 = 5 + 12 i$ $9+12i-4=5+12i$

What would $(3 + 2 i) (3 - 2 i) = ? ?$ $(3+2i)(3-2i)=??$

How do you multiply (3+2i)2(3+2i)^2?