How do you write the complex conjugate of the complex number #(6-3i)/ (2+i)#?

Question

1667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T09:43:32

The complex conjugate is #=9/5+12/5i#

If you want to simplify a quotient of complex numbers , multiply numerator and denominator by the conjugate of the denominator.

#z=z_1/z_2=(z_1barz_2)/(z_2barz_2)#

Also, #barz=barz_1/barz_2#

The conjugate of #(a+ib)# is #(a-ib)#

#i^2=-1#

So,

#(6-3i)/(2+i)=(3(2-i))/(2+i)#

#=(3(2-i)(2-i))/((2+i)(2-i))#

#=(3(4-4i+i^2))/(4-i^2)#

#=(3(3-4i))/(5)#

#=9/5-12/5i#

So, the complex conjugate is #=9/5+12/5i#

Second way,

The conjugate is #bar(6-3i)/bar(2+i)#

#=(6+3i)/(2-i)=((6+3i)(2+i))/((2-i)(2+i))#

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

#=(9+12i)/(5)#

#=9/5+(12i)/5#