Reciprocal of a complex number

**Morzilla** · January 15th 2008, 06:59 PM

I am asked to solve this

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

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

according to the book

$\frac{1}{3+4i}$ 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.

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

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

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

$\frac{3-4i}{9+16}$ = $\frac{3}{25}-\frac{4}{25}$ $i$

Where did the 25 come from?!

Thanks in advance to all!!

**topsquark** · January 15th 2008, 08:33 PM

Originally Posted by Morzilla

I am asked to solve this

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

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

according to the book

$\frac{1}{3+4i}$ 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.

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

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

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

$\frac{3-4i}{9+16}$ = $\frac{3}{25}-\frac{4}{25}$ $i$

Where did the 25 come from?!

Thanks in advance to all!!

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

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

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

-Dan

**Morzilla** · January 16th 2008, 02:50 AM

Originally Posted by topsquark

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

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

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

-Dan

and what about the 25 in the denominator, where did come from?!

**colby2152** · January 16th 2008, 04:56 AM

Originally Posted by Morzilla

and what about the 25 in the denominator, where did come from?!

$(3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$

**Morzilla** · January 16th 2008, 04:34 PM

Originally Posted by colby2152

$(3+4i)(3-4i) = 9 -16i^2 \Rightarrow 9 - (-16) = 9 + 16 \Rightarrow 25$

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

Much Thanks!!

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