How do you divide #i/(-2-8i)#?

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Answer 1 · 2016-11-24T07:24:54

#-2/17-1/34i#

You can multiply both the terms of the fraction by #-2+8i#:

#(i(-2+8i))/((-2-8i)(-2+8i))#

#(-2i+8i^2)/(4-64i^2)#

and, since #i^2=-1#,

#(-2i-8)/(4+64)#

#=(-2i-8)/68#

or, in the standard form:

#-8/68-(2i)/68=-2/17-1/34i#