Write in the form a+bi: (1-3i)/(-2-4i)

Not a kid with a home work problem, for a change... but an adult looking at Algebra II problem and wondering why he doesn't get their answer... obviously, I've forgotten how to do these. Show work please! Many thanks.

one way...

multiply numerator and denominator by conjugate.

notes: for two terms added/subtracted, the conjugate is same two terms with opposite sign between.

so conjugate of -2 - 4i is -2 + 4i

also, as long as we multiply top & bottom of fraction by same (non-zero) value, the fraction still represents original value.

so...

(1 - 3i)/(-2 - 4i)

= [(1 - 3i)(-2 + 4i)] / [(-2 - 4i)(-2 + 4i)]

now expand top & bottom (with FOIL, diff.sqs., etc)

= (-2 + 4i + 6i - 12i²) / (4 - 8i + 8i - 16i²)

= (-2 + 10i - 12i²) / (4 - 16i²)

recall that i² = -1 (the complex #'s), so replace i² with -1 to get

= (-2 + 10i + 12) / (4 + 16)

= (-10 + 10i) / 20

= -10/20 + 10i/20

= -1/2 + (1/2)i

or

= -0.5 + 0.5i


hope that helps

Oops! A sign got reversed!
The line that goes:
(-10 + 10i)/20

should be
(10 + 10i)/20

so the final answer is actually
1/2 + (1/2)i

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