# complex conjugate

Started by January 6, 2016
```Hello All,

Conceptually I understand complex conjugate but what is the application of
complex conjugates?

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```
```>Hello All,
>
>Conceptually I understand complex conjugate but what is the application
of
>complex conjugates?
>
>---------------------------------------
>Posted through http://www.DSPRelated.com

I can think of three main "uses":

1) To find the magnitude of a complex number.

z = a + bi
z_bar = a - bi

|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b )

2) To perform complex division:

z = (c+di)/(a+bi)

Multiply numerator and denominator by a-bi

z = (c+di)(a-bi)/(a*a+b*b)

3) The complex roots of polynomials with real valued coefficients come in
conjugate pairs.  The quadratic formula gives a clear example

Ax^2 + Bx + C = 0

x = -B/(2A) +/- sqrt(B^2-4AC)/(2A)

When the argument of the sqrt is negative this takes the form

x = c +/- di

It seems to me you meant to say you know the meaning "definitionally".
The mirror image with respect to the real axis comes to mind.

Ced

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```
```>>Hello All,
>>
>>Conceptually I understand complex conjugate but what is the application
>of
>>complex conjugates?
>>
>>---------------------------------------
>>Posted through http://www.DSPRelated.com
>
>I can think of three main "uses":
>
>1) To find the magnitude of a complex number.
>
>z = a + bi
>z_bar = a - bi
>
>|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b )
>
>2) To perform complex division:
>
>z = (c+di)/(a+bi)
>
>Multiply numerator and denominator by a-bi
>
>z = (c+di)(a-bi)/(a*a+b*b)
>
>3) The complex roots of polynomials with real valued coefficients come
in
>conjugate pairs.  The quadratic formula gives a clear example
>
>Ax^2 + Bx + C = 0
>
>x = -B/(2A) +/- sqrt(B^2-4AC)/(2A)
>
>When the argument of the sqrt is negative this takes the form
>
>x = c +/- di
>
>It seems to me you meant to say you know the meaning "definitionally".
>The mirror image with respect to the real axis comes to mind.
>

Dear Cedron,

Thanks. I see that in communication systems, receivers using matrix of
complex conjugate values (as sort of filters I guess). Hence my question.

A simple question on the first use-case,

>|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b )

if Z is in Cartesian form then one can directly compute sqrt (a*a + b*b).
In fact that is what is happening above. So, I am trying to understand how
introduction of complex conjugate term in the above equation helped?

---------------------------------------
Posted through http://www.DSPRelated.com
```
```>
>Dear Cedron,
>
>Thanks. I see that in communication systems, receivers using matrix of
>complex conjugate values (as sort of filters I guess). Hence my
question.
>
>A simple question on the first use-case,
>
>>|z| = sqrt( z * z_bar ) = sqrt( a*a + b*b )
>
>if Z is in Cartesian form then one can directly compute sqrt (a*a +
b*b).
>In fact that is what is happening above. So, I am trying to understand
how
>introduction of complex conjugate term in the above equation helped?
>
>---------------------------------------
>Posted through http://www.DSPRelated.com

Complex conjugates are used to find the magnitudes of complex vectors just
like with complex scalars.

When you're doing the math, as in this snippet I did in one of your other
threads, dealing with a complex number or vector as a single symbol is a
lot more convenient.

D^(*) * D = S^(*) * T^(*) * T * S

= S^(*) * I * S     ( I is the identity matrix )

= S^(*) * S

When you are implementing an algorithm in code it is generally more
efficient to use the component values.  On certain platforms, you have a
complex type so your implementation can resemble your math closer if that
is what you want.

a = ( z + z_bar ) / 2
b = ( z - z_bar ) / (2i)

These are the transition.  They are frequently found in the form:

cos( th ) = ( e^(i*th) + e^(-i*th) ) / 2
sin( th ) = ( e^(i*th) - e^(-i*th) ) / (2i)

Where 'th' stands for the Greek letter theta.

To understand why e^(i*th) is the complex conjugate of e^(-i*th), and to
understand complex numbers a little better, I recommend you read my first
blog article titled
"The Exponential Nature of the Complex Unit Circle" which can be found
here:

http://www.dsprelated.com/showarticle/754.php

Complex conjugates pop up in DFTs when you have a real valued signal.  The
top half (or negative half) of the DFT is the complex conjugate of the
bottom half.

This actually comes from the fact that the roots of unity for any N are