Division of complex numbers( include process)?

You need to multiply by the complex conjugate of the denominator:

Complex Conjugate:

(2-5i)* = 2 + 5i

Multiplying top and bottom:

(3+2i)*(2+5i) / { (2-5i)(2+5i) } =

(-4 + 19i) / 29

Multiply and divide by 2+5i

The denominator becomes = 2^2-(5i)^2 =4 -(25i^2)=4+25=29

(3-2i)(2-5i) /29

=(6-15i-4i+10i^2) / 29

=(6-19i-10)/29

=(-19i-4)/29

=-4/29 -19i/29

a+bi form a=-4/29 and b=-19/29

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