OK, so we saw something pretty interesting last time when we multiplied together complex numbers using the modulus argument form.

Remember that for two complex numbers which we will write as and , where are the moduli, and are the arguments of . If we multiply them together then we get:

Well, what would happen if the two complex numbers were the same? ie. if we have and we want ?

Well, then clearly:

.

What if we then multiplied this by one more time:

hmm, do we already see a pattern emerging? Let’s say that we have a complex number with modulus 1. Complex numbers of the form:

Are clearly modulus 1. We know that the modulus is the square root of the sum of the squares of the real and imaginary parts of a complex numbers so .

ok, so how about if we have where is an integer?

It turns out that there is a theorem, called de Moivre’s theorem which gives a very simple formula for this.

For integer . This is a particularly beautiful formula and the proof of it is by induction. However, we know that for induction we need a base case, and so when something is true for all integers (positive and negative) we must split it into two parts. First it is proved for , then it is proved for . You can find the full proof in the resource book, but you should try and work it out for yourselves first. How can we use this formula? Well, let’s say you want to calculate , you first need to calculate the modulus and argument of this complex number. We know that and we have that is an argument of this (remember that the argument is not unique. Then we have that:

However, we can see that and so because arguments are only unique up to integer addition of factors of we can subtract from 15.764 as many factors of as we want until the value of it is between and . In fact, and so this is a reasonable answer. We can see that the answer to our large power of a complex number is a number whose magnitude is and which lies at an angle just a little more than anticlockwise from the positive x-axis ie, it’s just under the negative axis. The answer is thus:

Let’s take an example which is slightly simpler to put into a graph. How about if we have for different . In the graph below we plot the first 100 . The first one, ie. is just above the real axis, and is red, and as increases the points go from red to green. The lines joining the points to the origin are just for reference. The complex numbers are themselves the points.

nkosinathiSeptember 3, 2015 at 10:26 amhi dr shock

this is a bit of a delayed reaction, above when we want to calculate the 17th power of the complex number and we reach the point where the angle is 15.764…. in a test situation will we be expected to know the value of 2.pi of will simply leaving in as (15.764…- (n)(2.pi)) be enough. let me know if my question is not so clear and ill try and rephrase.

Jonathan ShockSeptember 3, 2015 at 10:39 amHey there, you wouldn’t get a question like this in a test or exam. You might get a question where you end up with something like 17 pi and would be expected to know that if this is an argument then pi is also an argument, but getting the exact value with numerical coefficients wouldn’t be expected. I hope that makes sense.

All the best,

J

Mathemafrica - UCT MAM1000 lecture notes subject listOctober 4, 2015 at 10:28 am[…] MAM1000 lecture notes part 26 – Complex numbers – mod/arg form and multiplication […]

Azhar RohimanOctober 16, 2015 at 8:34 amDr. Shock, I think there is a mistake after you evaluated (3+4i)^17. You used an argument of 0.925 instead of 0.927.

RHMMUH005.

Jonathan ShockOctober 16, 2015 at 9:28 amWell spotted, yes!