How do you simplify # (2+3i) ^2#?

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Answer 1 · 2016-05-14T12:52:17

#-5 + 12i#

We can simply multiply this out like a binomial, remembering that #i^2=-1#.

#(2+3i)^2#

#implies (2+3i)(2+3i)#

#implies 4+6i+6i+9i^2#

#implies 4 + 12i -9#

#implies -5 + 12i#

NOTE this is different than the mod-squared of a complex number which is written #|z|^2 = z*barz# where #barz# is the complex conjugate of #z#.