How do you write #((10+4i)-(3-2i))/((6-7i)(1-2i))# in standard form?

1 Answer
Apr 21, 2018

Answer:

#(10+4i-(3-2i))/((6-7i)(1-2i))=-2/5+i/5#

Explanation:

#(10+4i-(3-2i))/((6-7i)(1-2i))#

Simplify numerator and multiply out denominator

#(7+6i)/(-8-19i)#

Multiply the numerator and denominator by the complex conjugate of the denominator.

#((7+6i)(-8+19i))/((-8-19i)(-8+19i))#

Apply the distributive property to the numerator and denominator.

#(-170+85i)/425#

Factor out the common factor.

#(85(-2+i))/(85(5))#

Simplify.

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