Reciprocal of a complex number

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• Jan 15th 2008, 06:59 PM
Morzilla
Reciprocal of a complex number
I am asked to solve this

$\displaystyle \frac{10}{3-4i}$

I looked over the book to see how this is done...

according to the book

$\displaystyle \frac{1}{3+4i}$ is to be written in stnrd form a+bi, so then one must multiply both the denominator and numerator by the opposite of the denominator in this case 3-4i.

$\displaystyle \frac{1}{3+4i}.\frac{3-4i}{3-4i}$=$\displaystyle \frac{3-4i}{9+16}$

Ok here is where I don't get it, what happened(s) to the $\displaystyle i$ and the signs in the denominator of $\displaystyle \frac{3-4i}{9+16}$

is the multiplication process of the denominator the same as using foil?! or am I just full of complex numbers!?

$\displaystyle \frac{3-4i}{9+16}$ = $\displaystyle \frac{3}{25}-\frac{4}{25}$$\displaystyle i Where did the 25 come from?! Thanks in advance to all!! (Handshake) • Jan 15th 2008, 08:33 PM topsquark Quote: Originally Posted by Morzilla I am asked to solve this \displaystyle \frac{10}{3-4i} I looked over the book to see how this is done... according to the book \displaystyle \frac{1}{3+4i} is to be written in stnrd form a+bi, so then one must multiply both the denominator and numerator by the opposite of the denominator in this case 3-4i. \displaystyle \frac{1}{3+4i}.\frac{3-4i}{3-4i}=\displaystyle \frac{3-4i}{9+16} Ok here is where I don't get it, what happened(s) to the \displaystyle i and the signs in the denominator of \displaystyle \frac{3-4i}{9+16} is the multiplication process of the denominator the same as using foil?! or am I just full of complex numbers!? \displaystyle \frac{3-4i}{9+16} = \displaystyle \frac{3}{25}-\frac{4}{25}$$\displaystyle i$

Where did the 25 come from?!

Thanks in advance to all!!

(Handshake)

Yes, you FOIL out the denominator as if "i" were a variable.

$\displaystyle (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 - 16i^2 = 9 + 16$

(Or you could just note that this is a $\displaystyle (a + b)(a - b) = a^2 - b^2$ form.)

-Dan
• Jan 16th 2008, 02:50 AM
Morzilla
Quote:

Originally Posted by topsquark
Yes, you FOIL out the denominator as if "i" were a variable.

$\displaystyle (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 - 16i^2 = 9 + 16$

(Or you could just note that this is a $\displaystyle (a + b)(a - b) = a^2 - b^2$ form.)

-Dan

and what about the 25 in the denominator, where did come from?!
• Jan 16th 2008, 04:56 AM
colby2152
Quote:

Originally Posted by Morzilla
and what about the 25 in the denominator, where did come from?!

$\displaystyle (3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$
• Jan 16th 2008, 04:34 PM
Morzilla
Quote:

Originally Posted by colby2152
$\displaystyle (3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$

yes, a math tutor showed me this in the math lab......boy I felt like such a Duh! :p

Much Thanks!!

(Handshake)