Show how the expression (-2i+1) is equivalent to (3-i)/(1+i)

= (3-i)x(1-i) / (1+i)x(1-i)

= (3i-3-i^2+1) / (i^2-1)

What is next

Thanks in advance

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- Jul 29th 2009, 03:39 PM05holtelrationalize a function
Show how the expression (-2i+1) is equivalent to (3-i)/(1+i)

= (3-i)x(1-i) / (1+i)x(1-i)

= (3i-3-i^2+1) / (i^2-1)

What is next

Thanks in advance - Jul 29th 2009, 03:54 PMKrizalid
We have

$\displaystyle 1-2i=\underbrace{1-2i\cdot \frac{1+i}{3-i}}_{1}\cdot \frac{3-i}{1+i},$

now $\displaystyle 1-2i\cdot \frac{1+i}{3-i}=\frac{1-i+2}{3-i}=1,$ and we're done. - Jul 29th 2009, 04:05 PM05holtel???
Well specifically the question asks "Which of the following expressions is equivalent to (3-i)/(1+i)" and then it gives you a choice of 4 expression (one being 1-2i .

Do I have to solve each one or is there a quicker way??? - Jul 29th 2009, 04:06 PM05holtel???
Also my friend said that you can plug (-1) into i^2 in my solution but I am not sure why

- Jul 29th 2009, 04:31 PMStroodle
$\displaystyle \frac{3-i}{1+i}$ $\displaystyle =\frac{(3-i)(1-i)}{(1+i)(1-i)}$ $\displaystyle =\frac{3-4i+i^2}{1-i^2}$ $\displaystyle =\frac{2-4i}{2}$ $\displaystyle =1-2i$

- Jul 29th 2009, 04:44 PMHallsofIvy
- Jul 29th 2009, 05:06 PM05holtel
I dont understand what i

^{2}= -1 means,

Sorry - Jul 29th 2009, 05:30 PMyeongil
HallsofIvy meant that $\displaystyle i^2 = -1$.

01 - Jul 29th 2009, 07:02 PMStroodle
Remember that $\displaystyle i=\sqrt{-1}$ so $\displaystyle i^2=(\sqrt{-1})^2=-1$

- Jul 30th 2009, 08:46 AM05holtel
Is "i" always equal to the square root of -1

Your patience is greatly appreciated [IMG]file:///C:/DOCUME%7E1/Elyse/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] - Jul 30th 2009, 09:16 AMHallsofIvy
I wince when I see that. -1, like any complex number except 0, has

**two**square roots. Unlike real numbers, since the complex numbers do not form and**ordered**field, we cannot just say "the positive number such that". I prefer just to assert that $\displaystyle i^2= -1$ and leave it at that. There are ways to define the complex numbers that avoid that problem.