How do you evaluate #(\frac { 1} { 1- 2i } + \frac { 3} { 1+ i } ) ( \frac { 3+ 4i } { 2- 4i } ) #?

1 Answer
Praveer Sharan
Nov 29, 2017

#(22+i)/(2-14i)#

Explanation:

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

Now, we take that answer and multiply it by #(3+4i)/(2-4i)#.

#((4-5i)/(3-i))((3+4i)/(2-4i)) = ((4-5i)(3+4i))/((3-i)(2-4i))#
#=(12 + 16i - 15i - 20i^2)/(6 - 12i - 2i + 4i^2)#
#=(12 + 16i - 15i + 20)/(6 - 12i - 2i - 4)# (since #i^2 = 1#)
#=(22 + i)/(2 - 14i)# (by simplifying)

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