SOLUTION: Solve in standard form (a + bi)
(2+i)/i
I first multiplied by i/i to get the following:
2i + i^2
=
2i - 1
=
-1 + 2i which is standard form so I thought I was
Question 489811: Solve in standard form (a + bi)
(2+i)/i
I first multiplied by i/i to get the following:
2i + i^2
=
2i - 1
=
-1 + 2i which is standard form so I thought I was correct.
However, the student solution manual that I have shows to multiply by -i/-i instead of i/i which is what I did. When doing so they get opposite answer that I do of 1 - 2i. I'm wondering why that manual suggests the answer is 1 - 2i and why they used the negative i over i versus positive.
I really just need to know or understand how they arrived at using the negative. Found 2 solutions by MathLover1, Edwin McCravy:Answer by MathLover1(20850) (Show Source):
It is perfectly OK to use . You will get the same answer
if you do it correctly. You just made a mistake.
The reason they tell you to use is because they
want ONE rule to work for all problems.
When you have
They always tell you to multiply by the conjugate of the denominator
over itself:
You problem is really the same as ,
so to follow that rule you would multiply by ,
which is the same as .
It would also work to use but they don't
want to tell you to multiply by the negative of the conjugate of
the denominator over itself.
Using
Since ,
Separate into two fractions:
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Or you can use
Since ,
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Edwin
