How do you simplify #(7i)/(8+i)#?

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Answer 1 · 2016-09-02T21:28:21

#(7i)/(8+i)=7/65+56/65i#

To simplify #(7i)/(8+i)#, we should multiply numerator and denominator by the complex conjugate of the denominator.

As complex conjugate of the denominator #(8+i)# is #(8-i)#, we have

#(7i)/(8+i)=(7i×(8-i))/((8+i)(8-i))#

= #(56i-7i^2)/(64+8i-8i-i^2)#

= #(56i-7×(-1))/(64+1)#

= #(7+56i)/65#

= #7/65+56/65i#

Answer 2 · 2016-09-04T14:32:45

#(7) / (65) + (56) / (65) i#

We have: #(7 i) / (8 + i)#

Let's multiply both the numerator and the denominator by the complex conjugate of the denominator:

#= (7 i) / (8 + i) cdot (8 - i) / (8 - i)#

#= ((7 i) (8) + (7 i) (- i)) / ((8)^(2) - (i)^(2))#

#= (56 i - 7 i^(2)) / (64 - i^(2))#

Let's apply the fact that #i^(2) = - 1#:

#= (56 i - (7 cdot (- 1))) / (64 - (- 1))#

#= (56 i + 7) / (65)#

#= (7) / (65) + (56) / (65) i#