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 2043 views around the world You can reuse this answer Creative Commons License