How do you simplify #(3+10i)^2# and write the complex number in standard form?

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Answer 1 · 2018-08-12T05:43:49

#-91 + 60i#

#(3+10i)^2#

#(3+10i)(3+10i)#

Use FOIL to distribute/expand:
#3 * 3 = 9#

#3 * 10i = 30i#

#10i * 3 = 30i#

#10i * 10i = 100i^2#

Combine them together:
#9 + 30i + 30i + 100i^2#

Combine like terms:
#9 + 60i + 100i^2#

We know that #i^2 = -1#, so:
#9 + 60i + 100(-1)#

#=9 + 60i - 100#

#-91 + 60i#

We know standard form is #a + bi# and this expression matches the form, so we are done.

Hope this helps!