How do you simplify # (2+5i)^2#?

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Answer 1 · 2016-01-27T17:09:08

#-21+20i#

First, distribute this using FOIL as you normally would a binomial.

#(2+5i)^2=(2+5i)(2+5i)=#

#overbrace(2xx2)^("First")+overbrace(2xx5i)^"Outside"+overbrace(5ixx2)^"Inside"+overbrace(5ixx5i)^"Last"#

This gives

#4+10i+10i+25i^2#

or

#4+20i+25i^2#

While this may look simplified, we can go one step further.

Since #i=sqrt(-1)#, we know that #i^2=-1#, so we can replace #i^2# with #-1#, giving:

#4+20i+25(-1)#

Now, subtract the #25# from #4#, giving our answer in the #a+bi# form of a complex number

#-21+20i#