SOLUTION: (2-4i)/(2) (2+4i)/(-5+2i) (1-i)/(-4-6i) (3+5i)/(-2+i)

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Question 252022: (2-4i)/(2)
(2+4i)/(-5+2i)
(1-i)/(-4-6i)
(3+5i)/(-2+i)

Found 2 solutions by jsmallt9, palanisamy:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The standard form for complex numbers is a + bi. So in each problem we want to end up with an expression of this form.

%282-4i%29%2F2
All we have to do here is split the fraction into two parts:
2%2F2+-+4i%2F2+
and simplify:
1+-+2i
and rewrite as an addition:
1+%2B+%28-2i%29

%282%2B4i%29%2F%28-5%2B2i%29
i+=+sqrt%28-1%29. So we can look at this problem as either division of complex numbers or as rationalizing the denominator of a fraction. Either way we do the same thing. Using the pattern %28a%2Bb%29%28a-b%29+=+a%5E2+-+b%5E2 we will turn the binomial (2-term expression) denominator into a binomial of perfect squares. Expressions of the form a+b and a-b are called conjugates. And this pattern tells us that if you multiply conjugates that you get the difference (subtraction) of the square of the first term and the square of the second term. This is a very handy way to eliminate square roots in binomials.

So we will multiply the numerator and denominator of
%282%2B4i%29%2F%28-5%2B2i%29
by the conjugate of the denominator: -5-2i
%28%282%2B4i%29%2F%28-5%2B2i%29%29%28%28-5-2i%29%2F%28-5-2i%29%29
Using FOIL on the top and FOIL or the pattern on the bottom we get:
%28-10+-+4i+-+20i+-8i%5E2%29%2F%28%28-5%29%5E2+-+%282i%29%5E2%29
which simplifies to:
%28-10+-+4i+-+20i+-8i%5E2%29%2F%2825+-+4i%5E2%29
Since i%5E2+=+-1:
%28-10+-+4i+-+20i+-8%28-1%29%29%2F%2825+-+4%28-1%29%29
%28-10+-+4i+-+20i+%2B8%29%2F%2825+%2B+4%29
Combining like terms:
%28-2+-+24i%29%2F29
Now we can split this into two terms:
-2%2F29+-+24i%2F29
In a + bi form:
-2%2F29+%2B+%28%28-24%29%2F29%29i

The last two problems
(1-i)/(-4-6i)
(3+5i)/(-2+i)
are just like this one. I'll leave them up to you. Just multiply the numerator and denominator by the conjugate of the denominator.

P.S. Using conjugates like this is also useful for rationalizing denominators like:
%282+%2B+sqrt%283%29%29%2F%285+-+sqrt%282%29%29


%2810+%2B+2sqrt%282%29+%2B+5sqrt%283%29+%2B+sqrt%286%29%29%2F%2825+-+2%29
%2810+%2B+2sqrt%282%29+%2B+5sqrt%283%29+%2B+sqrt%286%29%29%2F23
which may look like a mess but at least it's a mess with a rational denominator.

Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
(2-4i)/2 = 2(1-2i)/2 = 1-2i


(2+4i)/(-5+2i) = (2+4i)(-5-2i)/(-5+2i)(-5-2i)
= (-10-4i+20i+8)/[(-5)^2-(2i)^2]
= (-2+14i)/(25+4)
= 2(-1+7i)/29

(1-i)/(-4-6i) = (1-i)(-4+6i)/(-4-6i)(-4+6i)
= (-4+6i+4i+6)/[(-4)^2-(6i)^2]
= (2+10i)/(16+36)
= 2(1+5i)/52
= (1+5i)/26

(3+2i)/(-2+i) = (3+2i)(-2-i)/(-2+i)(-2-i)
= (-6-3i-4i+2)/ [(-2)^2-(i)^2]
= (-4-7i)/(4+1)
= -(4+7i)/5