# Complex Numbers 6: Dividing Complex Numbers

#### Nirvana

Joined Jan 18, 2005
58
What about dividing two complex numbers together, well before I get on to this you need to know about the complex Conjugate – what is it and why is it used?
Take two complex numbers in the form (a + jb) and divide one by the other;
Lets take (2 + j8) / (4 + j4) now look what we have here, we have a complex number on the bottom (denominator) of our fraction. This is undesirable for two reasons the first is that it looks rather messy and the second is that with a complex number as the denominator we can’t simplify the fraction. So our objective is to remove the complex number from thr bottom of the fraction and replace it with a real number. We need to do this without altering the value of our fraction. To explain what we are going to do take a simple fraction 2/3 now I can multiply this fraction by another fraction which will give me the value of one can’t I as this will just give us 2/3. Lets multiply 2/3 by 4/4 well the 4/4 gives 1 and 2/3 by 1 is 2/3. I could do the same by multiplying it by negative numbers as long as what I am multiplying by is 1, e.g. 2/3 multiplied by -9/-9, well again -9/-9 is one so we have 2/3.
This is what we have to do when we are faced with a complex number as the denominator of our fraction. Notice before though that I didn’t alter the value of our fraction as what I multiplied it by gave 1.
Now to remove the complex number from the bottom we have to multiply by what is called the complex conjugate of the complex number. The complex conjugate is exactly the same as the complex number we are dealing with except that we change the sign in the middle. So for example the complex conjugate of (3 + j6) is (3 – j6) and what we get out of it is a real number. Also the number what we get out is the difference of two squares that is 3 x 3 – 6 x 6 = 9 – 36 = -27

But we need to multiply the top and bottom of this fraction by the complex conjugate to give us one , as we did before so as not to alter the value of our fraction. All will become clear with this example. Lets divide the complex numbers (2 + j4) and (4 + j6)
[(2 + j4)/ (4 +j6)] x[(4 – j6) / (4 – j6)] it would be easier to write this down , I’ve used brackets to make it look easier. So first lets solve the top part of our fraction;
(2 + j4) x (4 – j6) = 8 – j12 + j16 – j squared 24 = 8 + j4 – j squared 24 ( as j squared is -1 we get) 6 + 24 + j4 = 30 + j4.

Now the bottom of our fraction (4 + j6) x (4 – j6) =
16 –j24 + j24 –j squared 36 = 16 – j squared 36 (as j squared is -1 we get) 16 + 36 = 52

So we have (30 + j4) as our numerator (top of fraction) and (52) as our Denominator (bottom of fraction).

So (30 + j4) / 52 = 0.58 + j0.08.

Note that we can divide the real and imaginary numbers by a real number.

Ok how about when the complex number is written in polar form, what then?

Well let the first complex number be r1(Cosθ1 + jSinθ1) and the second complex number be r2 (Cosθ2 + jSinθ2).

Then by dividing the first complex number by the second we get:
r1(Cosθ1 + jSinθ1)/ r2 (Cosθ2 + jSinθ2) , this is exactly the same as writing
r1/r2 [ (Cosθ1 + jSinθ1) / (Cosθ2 + jSinθ2)], look at what we have here, there is a complex number on the bottom of our fraction. We can’t have that can we, so we must multiply top and bottom of this fraction by the complex conjugate of the denominator which is (Cosθ2 - jSinθ2).

Now we have (excusing r1 and r2 for the moment) ;

[ (Cosθ1 + jSinθ1) / (Cosθ2 + jSinθ2)] x [(Cosθ2 - jSinθ2)/ (Cosθ2 - jSinθ2)]
Ok lets do the numerators first which are (Cosθ1 + jSinθ1) x (Cosθ2 - jSinθ2);
(Cosθ1 Cosθ2 –jSinθ2Cosθ1 + jSinθ1 Cosθ2 – j squared Sinθ1Sinθ2)
Remember that j squared is -1, we get the following.
Cosθ1 Cosθ2 –jSinθ2Cosθ1 + jSinθ1 Cosθ2 + Sinθ1Sinθ2.
Collect together real and imaginary components we get.
(Cosθ1 Cosθ2 + Sinθ1 Sinθ2) + j(Sinθ1Cosθ2 – Sinθ2 Cosθ1) – taken j out as common factor. Now remember your trig identities we get the following simplification.
Cos(θ1 –θ2) + jSin (θ1 – θ2) Remember this as our Numerator.
Lets tackle our Denominator.
(Cosθ2 + jSinθ2)/ (Cosθ2 - jSinθ2)
Cosθ2Cosθ2 –jSinθ2Cosθ2 + jSinθ2Cosθ2 – j squared Sinθ2Sinθ2
As j squared is -1 we get.
Cosθ2Cosθ2 –jSinθ2Cosθ2 + jSinθ2Cosθ2 + Sinθ2Sinθ2
Collecting real and imaginary parts together we get.
(Cosθ2Cosθ2 + Sinθ2Sinθ2) + j(Sin θ2Cosθ2 - Sin θ2Cosθ2)
(Cosθ2Cosθ2 + Sinθ2Sinθ2) + j(0)
(Cosθ2Cosθ2 + Sinθ2Sinθ2) – From our trig identity we know that Cos squared plus Sin squared is 1, so Our denominator is 1. Putting our fraction together we get;
[Cos(θ1 –θ2) + jSin (θ1 – θ2)]/1 = Cos(θ1 –θ2) + jSin (θ1 – θ2)
Now put our r’s back in and we get. r1/r2 (Cos(θ1 –θ2) + jSin (θ1 – θ2))
There we have it, to divide two complex numbers you simply divide the r’s together and take away the angles.
Here is an example, lets divide the complex numbers 10(Cos 60 + jSin80) and
2(Cos 20 + j Sin 30) and we get;
(10/2)(Cos(60 – 20) + jSin(80 – 30) = 5(Cos40 + jSin50)

Again we can have as many complex numbers as we wish being divided the result is always the same, simply divide the r’s and take away the angles.

#### Nirvana

Joined Jan 18, 2005
58
Complex numbers are used throughout engineering in many different branches. The most common use is when dealing with AC components, in particular the Capacitor, Inductor and Resistor - Commonly Known as Single Phase AC.

Well lets briefly recap on the relationships between voltage and current through each of these components.
In a resistor the current and voltage (Alternating Sinuosidally of course) are in phase with each other, what the voltage does the current does simultaneously (although their magnitudes may vary and usually do their positions match). In a Capacitor the current leads the voltage by 90 degrees and in an Inductor the current lags the voltage by 90 degrees. A useful word to remember is CIVIL and observe where the I (current ) and V (voltage) are with respect to C (Capacitor) and L (Inductor).

Now remember that 90 degrees was represented by the letter j and the -90 degrees was represented by -j.
Well lets see what this means then, we already know that the resistance of a Resistor is given by V/I and as I and V are in phase they are entirely real values (remember the meaning of the word real in complex numbers).

What about the resistance offered by the Capacitor which is known as Capacitive Reactance, well if a capacitor is used alone in a circuit the resistance would be its impedance where Z represents impedance so here Z = V/I now remember that V = IxXC (the XC is the capacitive reactance i.e. the resistance offered by the capacitor where XC is equal to 1/2πfC) so Z = V/I becomes Z = I XC / I (-90 degrees) as the voltage lags by 90 degrees. As -90 degrees is equal to -j the above becomes;
Z = V/I = (I XC / I) (-90 degrees) = -jXC
So basically what we have found out is that the Capacitive Reactance XC can be represented by -jXC.

What about the resistance offered by the Inductor, well if the inductor was used alone in a circuit we have the impedance Z to calculate. This impedance is equal to the total voltage V divided by the total current I so Z = V/I.
Now remembering that the voltage leads by 90 degrees and that we can represent this 90 degrees angle by the letter j we get; Z = V/I = (I x XL / I) (90 degrees) = jXL.
Here we have shown that we can represent the Inductive reactance XL by jXL. Although it might not seem obvious at this point why representing the quantities of XC and XL like this are important, it will be shortly.
NOTE: XL = 2πfL

In the above descriptions the quantities of -jXC and jXL can be represented in the form (a + jb) as a normal complex number. As XC and XL are just numbers indicating the magnitude of the resistance. -jXC can be written like (0 - jXC),
and XL can be written like (0 + jXL).

Nirvana.

#### Ahmed maqmdouh

Joined Jan 14, 2008
3
ty guys u are so great