How do you simplify #(3+i)/(1+4i)#?

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Answer 1 · 2016-12-03T20:07:08

There are several ways to do it. I like to multiply numerator and denominator by the complex conjugate of the denominator.

Multiply numerator and denominator by the complex conjugate of the denominator:

#(3 + i)/(1 + 4i)(1 - 4i)/(1 - 4i)#

Use the pattern, #(a + b)(a - b) = a^2 - b^2#, to multiply the denominator:

#((3 + i)(1 - 4i))/(1 - 16i^2)#

Use #i^2 = -1# to make the denominator a real number:

#((3 + i)(1 - 4i))/17#

Use the F.O.I.L. method to multiply the numerator.

#(3 - 12i + i - 4i^2)/17#

Use #i^2 = -1#:

#(3 - 12i + i + 4)/17#

Combine like terms:

#(7 - 11i)/17#

Distribute the 17:

#7/17 - 11/17i#