# Adding and Subtracting Complex Numbers

• Oct 14th 2008, 08:23 AM
EyesForEars
Adding and Subtracting Complex Numbers
Ok. I understand that a complex number can be written in the form...a+bi.
Which leads me to the idea that
(5-6i)-4i is equivalent to (5-6i)-(0-4i).
Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So..
(5-6i)-4i = 5-2i
Is this a true statement?
My text says no, and I can't wrap my head around it.

P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i
• Oct 14th 2008, 09:31 AM
masters
Quote:

Originally Posted by EyesForEars
Ok. I understand that a complex number can be written in the form...a+bi.
Which leads me to the idea that
(5-6i)-4i is equivalent to (5-6i)-(0-4i).

The above should be written (5-6i)-4i = (5-6i) - (0 + 4i). Remember, when you put grouping symbols around an expression and precede the group with a negative sign, each sign within the group must be changed to its opposite.

-4i = -(+4i)

Which gives me 5-2i as a solution, because 5-0=5 and -6i-(-4i) is -2i. So.. (5-6i)-4i = 5-2i

Removing the parentheses on the above we get 5 - 6i - 4i = 5 - 10i

Is this a true statement?
My text says no, and I can't wrap my head around it.

P.S. (a+bi)-(c+di)= a+bi-c-di = (a-c)+(b-d)i

Using this, we get (5-6i)-(0+4i) = 5-6i-0-4i = (5-0)+(-6-4)i = 5-10i

..
• Oct 14th 2008, 10:06 AM
EyesForEars
So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.

I think I got it.

So, I'll say.

(19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i
• Oct 14th 2008, 10:24 AM
masters
Quote:

Originally Posted by EyesForEars
So, any time I apply grouping symbols to an expression preceeded by a negative sign, I have to then reverse the signs of said terms. Got it.

I think I got it.

So, I'll say.

(19-26i)-7i = (19-26i)-(0+7i) = 19-26i-0-7i = 19-33i

You are correct, my friend!!(Clapping)
• Oct 14th 2008, 03:42 PM
djmccabie
dont forget "--" = +

:)