Complex Numbers

  • Thread starter Master J
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  • #1
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Hey guys.

So, complex numbers.

Its been a while since I've dealt with them. I've been wondering a few things about them.

When expressed in the form a + ib, is there a particular way that you square them?

And when you have a rational number, with an i term on the bottom, how exactly do you go about simplifying it into non rational form?


Cheers guys. I'm fairly rusty when it comes to the complex numbers. Hope this will clear some things up for me. I need to be fairly sharp at them now that I'm doing Physical Optics.
 

Answers and Replies

  • #2
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You can treat the i like another variable. When you multiply two complex numbers, let's say a+ib and c+id, it is just like multiplying two binomials:

(a+ib)(c+id) = ac + aid + ibc + i2bd = (ac-bd) + i(bc+ad).

As for division (in the case of your rational number), what you want to do is to make the denominator real by multiplying top and bottom by the complex conjugate. For example, suppose you have (a+ib)/(c+id). The complex conjugate of c+id is c-id (the sign of the imaginary part is switched). When you multiply top and bottom by c-id, you will get:

[(ac+bd) + i(bc-ad)] / (c2 + d2)

The denominator is real, so you basically just have the solution as a complex number.
 
  • #3
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Cheers for the help. To make sure I have this, could someone perhaps show hoe this is seperated into the form a + i b:


u - 1 = [N(e^2)] / [vm(g^2 - w^2 -iyw)]

Real: 1 + N(e^2)(g^2 - w^2) / [vm[(g^2 - w^2)^2 + (yw)^2]]


Imaginary: N(e^2)yw / [vm[(g^2 - w^2)^2 + (yw)^2]]
 

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