How do you simplify (6+i)/(6-i)?

1 Answer
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