How do you simplify #(–3i)/(5+4i)#?

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Answer 1 · 2015-12-02T10:56:39

#(-12-15i)/(41)#

Given: #(-3i)/(5+4i)#

Multiply by #1 " in the form of " (5-4i)/(5-4i)# giving:

#((-3i)(5-4i))/((5+4i)(5-4i))#

#(-15i+12i^2)/( (25-20i+20i-16i^2)#

But #i^2=-1# giving:

#(-12-15i)/(25+16) #

#(-12-15i)/(41)#