How do you simplify #(6+i)/(6-i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Lucio Falabella Jan 25, 2016 #z=(6+i)/(6-i)=(35+12i)/37=35/37+12/37i# Explanation: Given: #z=z_1/z_2# with #z_1,z_2 in CC# to semplify #z# you have to moltiplicate both numerator and denominator by #bar(z_2)#, the complex conjugate of #z_2# #:.# if #z_2=a+ib=>bar(z_2)=a-ib# #:. z=(6+i)/(6-i)*(6+i)/(6+i)=(6+i)^2/(36-i^2)# Remebering that #i=sqrt(-1)=>i^2=-1# #(6+i)^2/(36-i^2)=(36+12i+i^2)/(36-(-1))=(36-1+12i)/37=(35+12i)/37=# #=35/37+12/37i# Answer link Related questions How do I graphically divide complex numbers? How do I divide complex numbers in standard form? How do I find the quotient of two complex numbers in polar form? How do I find the quotient #(-5+i)/(-7+i)#? How do I find the quotient of two complex numbers in standard form? What is the complex conjugate of a complex number? How do I find the complex conjugate of #12/(5i)#? How do I rationalize the denominator of a complex quotient? How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#? How do you write #(-2i) / (4-2i)# in the "a+bi" form? See all questions in Division of Complex Numbers Impact of this question 1867 views around the world You can reuse this answer Creative Commons License