How do you divide #20 / (3+i)#?

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Answer 1 · 2018-03-16T16:02:47

#6-2i#

Let's turn #i# to a square root.

#=>20/(3+sqrt(-1))# Let's multiply this by the conjugate of #3+sqrt(-1)#

Remember that the conjugate of #a+bi# is #a-bi#

We have:

#=>20/(3+sqrt(-1))*(3-sqrt(-1))/(3-sqrt(-1))#

#=>(20(3-sqrt(-1)))/((3-sqrt(-1))(3+sqrt(-1)))#

Expand this to get:

#=>(60-20sqrt(-1))/(9+3sqrt(-1)-3sqrt(-1)-(sqrt(-1))^2)#

#=>(60-20sqrt(-1))/(9-(sqrt(-1))^2)#

#=>(60-20sqrt(-1))/(9-(-1))#

#=>(60-20sqrt(-1))/(10)# Simplify to get:

#=>6-2sqrt(-1)#

#=>6-2i#