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

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Answer 1 · 2016-01-25T23:04:11

#z=(6+i)/(6-i)=(35+12i)/37=35/37+12/37i#

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#