# rationalize a function

• Jul 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

• Jul 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.
• Jul 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???
• Jul 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
• Jul 29th 2009, 04:31 PM
Stroodle
$\frac{3-i}{1+i}$ $=\frac{(3-i)(1-i)}{(1+i)(1-i)}$ $=\frac{3-4i+i^2}{1-i^2}$ $=\frac{2-4i}{2}$ $=1-2i$
• Jul 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 $i^2= -1$.
• Jul 29th 2009, 05:06 PM
05holtel
I dont understand what i2= -1 means,
Sorry
• Jul 29th 2009, 05:30 PM
yeongil
HallsofIvy meant that $i^2 = -1$.

01
• Jul 29th 2009, 07:02 PM
Stroodle
Remember that $i=\sqrt{-1}$ so $i^2=(\sqrt{-1})^2=-1$
• Jul 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]
• Jul 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.