How do you divide #(4+5i)/(2-3i)#?

2 Answers
Jan 10, 2016

Answer:

Multiply both numerator and denominator by the Complex conjugate of the denominator, then simplify to find:

#(4+5i)/(2-3i)=-7/13 + 22/13i#

Explanation:

Multiply numerator and denominator by #2+3i# (using FOIL if it helps) and simplify:

#(4+5i)/(2-3i)#

#=((4+5i)(2+3i))/((2-3i)(2+3i))#

#=(stackrel "First" overbrace((4*2))+stackrel "Outside" overbrace((4*3i))+stackrel "Inside" overbrace((5i*2))+stackrel "Last" overbrace((5i*3i)))/(2^2+3^2)#

#=(8+12i+10i-15)/(4+9)#

#=(-7+22i)/13#

#=-7/13 + 22/13i#

Jan 10, 2016

Answer:

#=-7/13 + 22/13i#

Explanation:

#(4+5i)/(2-3i)#

Whenever we divide complex numbers we multiply both numerator and denominator with the complex conjugate of the denominator, this makes the denominator a real number

If the complex number is #a+ib# then the complex conjugate is #a-ib#

For example
#(a+ib)(a-ib)#
# = (a)^2 - (ib)^2#
#=a^2-i^2b^2#
#=a^2-(-1)b^2#
#=a^2+b^2# this is a real number.

Now back to our problem.

#(4+5i)/(2-3i)#

#=(4+5i)/(2-3i)*(2+3i)/(2+3i)#

#=((4+5i)(2+3i))/(2^2+3^2)#

#=(4(2)+4(3i)+5i(2) + 5i(3i))/(4+9)#

#=(8+12i+10i+15i^2)/13#

#=(8+22i+15(-1))/13#

#=(8-15+22i)/13#

#=(-7+22i)/13# Answer

#=-7/13 + 22/13i quad# Answer in #a+ib# form