How do you rationalize #6/(3-3i)#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer Nghi N. May 12, 2015 Multiply both numerator and denominator by (3 + 3i) #[6(3 - 3i)]/(9 - 3) #= #[18(1 - i)]/6# =# 3(1 - i)# Answer link Related questions What is a complex conjugate? How do I find a complex conjugate? What is the conjugate zeros theorem? How do I use the conjugate zeros theorem? What is the conjugate pair theorem? How do I find the complex conjugate of #10+6i#? How do I find the complex conjugate of #14+12i#? What is the complex conjugate for the number #7-3i#? What is the complex conjugate of #3i+4#? What is the complex conjugate of #a-bi#? See all questions in Complex Conjugate Zeros Impact of this question 1733 views around the world You can reuse this answer Creative Commons License