How do you divide #(2-4i)/(4-3i)#?

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Answer 1 · 2016-01-27T02:11:26

#4/5 -2/5 i#

Remember: #i^2 = -1#

Given #(2-4i)/(4-3i)#

We can simplify this expression by multiply by the conjugate of the denominator

#((2-4i)/(4-3i))color(red)(( (4+3i)/(4+3i))#

Expand the expression, then combine like terms,

#(8+6i -16i -12i^2) /(16-12i +12i -9i^2)#

#= (8 -16i -12(-1))/(16-9(-1))#

#= (20 -10i)/(25)#

#= (20)/25 -(10)/25 i#

#=4/5 -2/5 i#