How do you simplify (6-3i)/ (2+i)6−3i2+i?

1 Answer
Gió
Dec 1, 2015

I found: 9/5-12/5i95−125i

Explanation:

You need to change the denominator into a pure real and then divide. To do this you can multiply and divide by the complex conjugate of the denominator 2-i2−i to get:

(6-3i)/(2+i)*(2-i)/(2-i)=6−3i2+i⋅2−i2−i=
=(12-6i-6i+3i^2)/(4+1)==12−6i−6i+3i24+1= remembering that i^2=-1i2=−1
=(12-12i-3)/5=9/5-12/5i=12−12i−35=95−125i

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