How do you simplify #(3+9i)/(3i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Ratnaker Mehta Sep 5, 2016 #(3+9i)/(3i)=(cancel(3)(1+3i))/((cancel3)i)=(1+3i)/i=(1+3i)/i xxi/i=(i+3i^2)/i^2=(i-3)/-1=3-i#. Explanation: Alternatively, #(3+9i)/(3i)=3/(3i)+(9i)/(3i)=1/i+3={(1*i)/(i*i)}+3# #=i/i^2+3=i/-1+3=3-i#. Answer link Related questions How do I graphically divide complex numbers? How do I divide complex numbers in standard form? How do I find the quotient of two complex numbers in polar form? How do I find the quotient #(-5+i)/(-7+i)#? How do I find the quotient of two complex numbers in standard form? What is the complex conjugate of a complex number? How do I find the complex conjugate of #12/(5i)#? How do I rationalize the denominator of a complex quotient? How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#? How do you write #(-2i) / (4-2i)# in the "a+bi" form? See all questions in Division of Complex Numbers Impact of this question 2368 views around the world You can reuse this answer Creative Commons License