How do you divide #(-5-9i)/(9+8i)#?

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Answer 1 · 2016-11-13T09:18:59

#13/5-41/145i#

You can multiply the terms of the fraction by the same expression:

#9-8i#

to get a denominator without i:

#((-5-9i)(9-8i))/((9+8i)(9-8i))#

#=(-45+40i-81i+72i^2)/(81-64i^2)#

#=(-45-41i-72)/(81+64)# (since #i^2=-1#)

#(-117-41i)/145#

and, in the standard form:

#-117/45-41/145i#

#=13/5-41/145i#