Tuesday, March 9, 2010

More on Complex Numbers

We left off in the previous instalment obtaining the vector C in complex form, along with its argument.

Now we look at the vectors A and B, which will call henceforth z1 and z2 to be consistent with complex notation. Our eventual goal will be to find the resultant- which will come in the next instalment. In the meantime we will be working toward showing the multiplication and division of two complex forms, call them z1 and z2:

[z1 + z2]

From the diagram (see previous blog):

A= z1 = -2 + 2i

B = z2 = -2 -3i

So:

z1 = x1 + iy1

arg(z1) = arctan(y1/x1) = arctan (-2/2) = arctan(-1)

so (theta_1) = -45 degrees

Now find r1:

r1 =[x1^2 + y1^2]^1/2 = [1^2 + 1^2]^1/2 = [2]^1/2 = 1.414

Therefore:

z1 = 1.414 (cos(-45) + isin(-45)) = 1.414 cis(-45)

We now turn to the vector B which is:

z2 = x2 + iy2= -2 -3i

then: arg(z2) = arctan(y2/x2) = arctan (-3/-2) = arctan (3/2) = 56.3 deg

While:

r2 =[x2^2 + y2^2]^1/2 = [(-2)^2 + (-3)^2]^1/2 = [13]^1/2 = 3.6

Therefore:

z2 = 3.6(cos(56.3) + isin(56.3) = 3.6 cis(56.3)

Now, how do we obtain the complex product: [z1*z2]?

We have that:

[z1*z2] = (z1*z2) cis(arg(z1) – arg(z2))

But:

(z1*z2) = (1.414)x (3.6) = 5.1

And:

arg(z1) – arg(z2) = (-45) – (56.3) = -101.3

so that:

[z1*z2] = 5.1 cis(-101.3) = 5.1 (cos (-101.3) + isin(-101.3))

[z1*z2] = 5.1((-0.195) + i(-0.98))

[z1*z2]= 0.99 + 0.98i

Next Series topic: Complex division and complex resultants