How do you write #((5-2i)+(5+3i))/((1+i)-(2-4i))# in standard form?

1 Answer
marfre
Feb 18, 2017

#-5/26 -51/26 i#

Explanation:

Add like terms in the numerator and denominator:
#((5+5) + (-2i +3i))/((1+ -2)+(i--4i)) = (10+i)/(-1+5i)#

Multiple by #1# using the conjugate of the denominator:
:#(10+i)/(-1+5i) * (-1-5i)/(-1-5i) = (-10-50i-i-5i^2)/(1-25i^2) #

Remember that #i^2 = -1#:

#(-10-50i-i+5)/(1+25) = (-5-51i)/(26)#

Put in standard form: #-5/26 -51/26 i#

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