How do you write #(2+3i)^2# in standard form?

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Answer 1 · 2015-12-14T18:09:17

#(2+3i)^2 = -5 + 12 i #

Rewrite the expression as

#(2+3i)^2 = (2+3i)(2+3i)#

Expand by multiplying the expression to get

#(4+ 6i +6i +9i^2)#

Combine like terms

#(4 + 12 i + 9i^2)#

Replace #i^1= -1# , this is be definition if imaginary number

#(4 + 12 i + 9 *(-1)) => 4 - 9 + 12i => -5 + 12i #

#(2+3i)^2 = -5 + 12 i #