How do you simplify #(15-8i)^2# and write the complex number in standard form?

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Answer 1 · 2016-07-28T15:41:41

#161-240i#

Since
#(a+b)^2=a^2+2ab+b^2#,

then

#(15-8i)^2=15^2+2(15)(-8i)+(-8i)^2#

that is

#225-240i+64i^2#

and, since #i^2=-1#, it becomes:

#225-240i-64=161-240i# in standard form