# Math Help - rationalize a function

1. ## 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

2. 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.

3. ## ???

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???

4. ## ???

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

5. $\frac{3-i}{1+i}$ $=\frac{(3-i)(1-i)}{(1+i)(1-i)}$ $=\frac{3-4i+i^2}{1-i^2}$ $=\frac{2-4i}{2}$ $=1-2i$

6. 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 $i^2= -1$.

7. I dont understand what i2= -1 means,
Sorry

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

01

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

10. 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]

11. 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.