# Complex Math Question

• posted

Is this correct...

I have a complex expression Numerator/Denominator where Numerator is the complex conjugate of the Denominator.

Am I correct that the net phase angle is twice that of the Numerator?

That would simplify my life if I didn't have to make the Denominator real, which turns into an Algebraic mess :-( ...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

That doesn't seem right. if N = A+iB then D = A-iB if I multiply top and bottom by A+iB I get (A+iB)^2/(A^2+B^2)

(Unless I didn't understand)

George H

• posted

Sure. Just convert to polar coordinates and it's clear.

A exp(j phi)

----------- == exp(j 2 phi) A exp(-j phi)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal Consultant ```
• posted

Duh, (Scuffs shoe in dirt, hangs head in shame.) George H.

• posted

Don't feel bad. I just proved it to myself by backing up to a sane expression....

(A-jB)/(A+jB)

and checking against the TAN(2x) expression.

I got myself balled-up with a messy complex expression (fitting an all-pass to specific angles).

...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

Well, the net phase angle _is_ twice the numerator's.

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Tim Wescott
Wescott Design Services ```
• posted

If it was simple, they wouldn't have called it "complex".

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Tim Wescott
Wescott Design Services ```
• posted

:-D ...Jim Thompson

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| James E.Thompson                                 |    mens     |
• posted

Hello? Conjugate is just the number rotated 180o on the Argand diagram, angles subtract upon division, so quotient has angle twice the numerator. We don't need no stinking algebra to figure that one out.

• posted

I assume you mean Complex as in sqrt(-1), not Complex as in difficult. Rotating, charged colliding black holes are difficult, sums are not...

Kevin Aylward B.Sc.

- SuperSpice

• posted

Yes

:-D

...Jim Thompson

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| James E.Thompson                                 |    mens     |