How do you simplify #(-3 + 2i) / (2 - 5i)#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer Shwetank Mauria Jun 30, 2016 #(-3+2i)/(2-5i)=-16/29-11/29i# Explanation: To simplify #(-3+2i)/(2-5i)#, we need to multiply numerator and denominator by complex conjugate of the denominator i.e. here #2+5i#. Hence, #(-3+2i)/(2-5i)=((-3+2i)(2+5i))/((2-5i)(2+5i))# = #(-6-15i+4i+10i^2)/(4-25i^2)# = #(-6-15i+4i-10)/(4+25)# = #(-16-11i)/29=-16/29-11/29i # 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 6191 views around the world You can reuse this answer Creative Commons License