How do you divide #(5i)/(-2-6i)#?

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Answer 1 · 2016-11-13T17:32:49

#(5i)/(-2-6i)=3/4-1/4i#

Whenever you divide a complex number by another complex number, you write it in fractional form

and then multiply numerator and denominator by complex conjugate of denominator

Complex conjugate of #a+bi# is #a-bi#. Hence that of #-2-6i# is #-2+6i#

So #(5i)/(-2-6i)xx(-2+6i)/(-2+6i)#

= #(-10i-30i^2)/((-2)^2-(6i)^2)#

as #i^2=-1#, this becomes

#(-10i+30)/(4+36)=(30-10i)/40=3/4-1/4i#