Dividing complex numbers...

[losfomot]

I am sure this is very simple, but could someone work through this question for me? This is not homework, I just want to learn the steps to do it.

 

x = (4-i)/(1-3i)

 

I am supposed to put it into the form x = a + bi

 

The answer is:

 

x = 7/10 - 11/10 i

 

Thanks for any help you can give me.

[elfstone]

You should multiply each part with 1+3i. Then it would be :

x = (4-i)(1+3i)/1^2-(3i)^2 --> x = (7-11i)/10 which is the answer you seek.

 

The bottom part is because of a^2 - b^2 = (a+b)*(a-b)

[losfomot]

You should multiply each part with 1+3i. Then it would be :

x = (4-i)(1+3i)/1^2-(3i)^2 --> x = (7-11i)/10 which is the answer you seek.

 

The bottom part is because of a^2 - b^2 = (a+b)*(a-b)

 

 

Thanks A lot!

 

Here's another one that has to be put into the same form:

 

x = -1/(4i)

 

Any help appreciated.

[jordan]

Following the previous post's logic, you want to get rid of the complex denominator. I'd just multiply top and bottom by 4i. The only tricky part about this one is that the wont be a real number part to the solution. In other words, the wont be an "a", just a "bi" part.

[losfomot]

Following the previous post's logic, you want to get rid of the complex denominator. I'd just multiply top and bottom by 4i. The only tricky part about this one is that the wont be a real number part to the solution. In other words, the wont be an "a", just a "bi" part.

 

 

Duh.

 

so the answer is 1/4 i

 

Thanks.

[jordan]

Sure thing.

[BobbyJoeCool]

ah, but there is a real part...

 

EVERY number has a real and a non-real part, in this case a=0.

 

so 0+(1/4)i=x

 

whereas the common number 7 is really 7=7+0i.

 

completely irrellevent, but interesting fact anyway.

[jordan]

True, but I meant that, just as we don't bother to write 0i for every real number, we don't have to bother to write 0 for this answer. Know what you mean though. Good to point it out.