# Imaginary numbers

• Mar 5th 2010, 11:12 PM
demode
Imaginary numbers
In the following, I don't understand how the expression on the LHS equals to the one on the RHS:

$Im \left( \frac{(-e^{\pi}-1)(1-i)}{(1+i)(1-i)} \right) =\frac{e^{\pi}+1}{2}$

("Im" means "imaginary")

Of course if on LHS we cancel out the (1-i) term on the numerator with the one on denominator we get

$Im \left( \frac{(-e^{\pi}-1)}{(1+i)} \right)$

And if $i=1$, then the denominator becomes 2. But I can't work out the rest. Any help is appreciated.
• Mar 6th 2010, 02:09 AM
Prove It
Quote:

Originally Posted by demode
In the following, I don't understand how the expression on the LHS equals to the one on the RHS:

$Im \left( \frac{(-e^{\pi}-1)(1-i)}{(1+i)(1-i)} \right) =\frac{e^{\pi}+1}{2}$

("Im" means "imaginary")

Of course if on LHS we cancel out the (1-i) term on the numerator with the one on denominator we get

$Im \left( \frac{(-e^{\pi}-1)}{(1+i)} \right)$

And if $i=1$, then the denominator becomes 2. But I can't work out the rest. Any help is appreciated.

You need to convert the whole complex number into the form $a + ib$ so that you can read off the imaginary part.

$\frac{(-e^{\pi} - 1)(1 - i)}{(1 + i)(1 - i)} = \frac{-e^{\pi} + e^{\pi}i - 1 + i}{1 + 1}$

$= \frac{-e^{\pi} - 1 + (e^{\pi} + 1)i}{2}$

$= \frac{-e^{\pi} - 1}{2} + \left(\frac{e^{\pi} + 1}{2}\right)i$.

So what is the imaginary part of this complex number?
• Mar 6th 2010, 02:54 AM
HallsofIvy
Quote:

Originally Posted by demode
In the following, I don't understand how the expression on the LHS equals to the one on the RHS:

$Im \left( \frac{(-e^{\pi}-1)(1-i)}{(1+i)(1-i)} \right) =\frac{e^{\pi}+1}{2}$

("Im" means "imaginary")

Of course if on LHS we cancel out the (1-i) term on the numerator with the one on denominator we get

$Im \left( \frac{(-e^{\pi}-1)}{(1+i)} \right)$

And if $i=1$, then the denominator becomes 2. But I can't work out the rest. Any help is appreciated.

What do you mean "if i= 2". Do you solve problems with real numbers by saying "if 1= 2"? You seem to think that "i" is a variable- it is not!

If you are going to do problems with imaginary and complex numbers, it would be a really good idea to first learn what "imaginary" and "complex numbers" are.