How do you rationalize the denominator of #(sqrt-9)/((4-7i)-(6-6i))#?

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Answer 1 · 2017-03-11T18:15:21

#(sqrt-9)/((4-7i)-(6-6i))=-3/5-6/5i#

#(sqrt-9)/((4-7i)-(6-6i))#

= #(sqrt((3i)^2))/((4-7i-6+6i)#

= #(3i)/(-2-i)#

If we have a complex number #a+bi# as denominator,

to rationalize it we multiply numerator and denominator by its complex conjugate i.e. #a-bi# as it makes the denominator #a^2+b^2#.

Hence in the given case as denominator is #-2-i#, we should multiply numerator and denominator by #-2+i#, which gives us

#((3i)(-2+i))/((-2-i)(-2+i))#

= #(-6i+3i^2)/((-2)^2+1^2)#

= #(-3-6i)/5#

= #-3/5-6/5i#