rationalize a function

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July 29th 2009, 03:39 PM
05holtel

rationalize 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
July 29th 2009, 03:54 PM
Krizalid

We have

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

now $1-2i\cdot \frac{1+i}{3-i}=\frac{1-i+2}{3-i}=1,$ and we're done.
July 29th 2009, 04:05 PM
05holtel

???

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???
July 29th 2009, 04:06 PM
05holtel

???

Also my friend said that you can plug (-1) into i^2 in my solution but I am not sure why
July 29th 2009, 04:31 PM
Stroodle

$\frac{3-i}{1+i}$ $=\frac{(3-i)(1-i)}{(1+i)(1-i)}$ $=\frac{3-4i+i^2}{1-i^2}$ $=\frac{2-4i}{2}$ $=1-2i$
July 29th 2009, 04:44 PM
HallsofIvy

Quote:

Originally Posted by 05holtel

Also my friend said that you can plug (-1) into i^2 in my solution but I am not sure why

Because i²= -1, of course.
Sorry. "sup" is "HTML" for "superscript". It should be $i^2= -1$ .
July 29th 2009, 05:06 PM
05holtel

I dont understand what i²= -1 means,
Sorry
July 29th 2009, 05:30 PM
yeongil

HallsofIvy meant that $i^2 = -1$ .

01
July 29th 2009, 07:02 PM
Stroodle

Remember that $i=\sqrt{-1}$ so $i^2=(\sqrt{-1})^2=-1$
July 30th 2009, 08:46 AM
05holtel

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]
July 30th 2009, 09:16 AM
HallsofIvy

Quote:

Originally Posted by Stroodle

Remember that $i=\sqrt{-1}$ so $i^2=(\sqrt{-1})^2=-1$

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 $i^2= -1$ and leave it at that. There are ways to define the complex numbers that avoid that problem.

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