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

2 Answers
Jan 10, 2016

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

=-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