**.**

*Graphical solution for the resultant, z5 - can be compared with algebraic result*Back to our favorite numbers! We left off with a couple of problems which we’ll now look at. I asked readers to try the division:

(1 + i) by [3]^1/2 – i

The first business is to get each into polar form, specifically as a (cis) function:

Then (1 + i) = z1 = x1 + iy1, so arg(z1) = arctan (y1/x1)

Further:

arctan (y1/x1) = arctan (1/1) = arctan (1) so theta1 = 45 deg

What about r1?

r1= [1^2 + 1^2]^1/2 = [2]^1/2 = 1.4

so z1 = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)

Now: z2 = [3]^1/2 – i

So arg(z2) = arctan(y2/x2) = arctan(-1/ [3]^1/2) so theta2 = (-30 deg)

And for r2: r2 = [([3]^1/2)^2 + (-1)^2]^1/2 = [4]^1/2 = 2

Then: z2 = 2[cos(-30) +isin(-30)] = 2cis(-30)

We divide:

(z1/z2)

Which means dividing the r’s first: r1/r2 = 1.4/ 2 = 0.707

Then subtract angles: [(theta1 – theta2)] = {(45 deg) – (-30 deg)} = 75 degrees

So the end result of the division is:

(z1/z2) = 0.707 cis(75) = 0.707{cos(75) + isin(75)}

Since cos(75) = 0.258 and sin(75) =0.966, we have:

(z1/z2) = 0.707[(0.258) + i(0.966)] = 0.183 + 0.683i

Now there was also another question inserted into the content earlier. Recall:

z5 = [z3 + z4] = [(-4 – i) + (4 + 3i)] = 0 + 2i.

(1 + i) by [3]^1/2 – i

The first business is to get each into polar form, specifically as a (cis) function:

Then (1 + i) = z1 = x1 + iy1, so arg(z1) = arctan (y1/x1)

Further:

arctan (y1/x1) = arctan (1/1) = arctan (1) so theta1 = 45 deg

What about r1?

r1= [1^2 + 1^2]^1/2 = [2]^1/2 = 1.4

so z1 = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)

Now: z2 = [3]^1/2 – i

So arg(z2) = arctan(y2/x2) = arctan(-1/ [3]^1/2) so theta2 = (-30 deg)

And for r2: r2 = [([3]^1/2)^2 + (-1)^2]^1/2 = [4]^1/2 = 2

Then: z2 = 2[cos(-30) +isin(-30)] = 2cis(-30)

We divide:

(z1/z2)

Which means dividing the r’s first: r1/r2 = 1.4/ 2 = 0.707

Then subtract angles: [(theta1 – theta2)] = {(45 deg) – (-30 deg)} = 75 degrees

So the end result of the division is:

(z1/z2) = 0.707 cis(75) = 0.707{cos(75) + isin(75)}

Since cos(75) = 0.258 and sin(75) =0.966, we have:

(z1/z2) = 0.707[(0.258) + i(0.966)] = 0.183 + 0.683i

Now there was also another question inserted into the content earlier. Recall:

z5 = [z3 + z4] = [(-4 – i) + (4 + 3i)] = 0 + 2i.

WHERE would this resultant be and what is arg(z5)?

Here, the resultant is first shown in the diagram, via graphical completion of the parallelogram formed by the vectors z3, and z4.. Readers can check to see it is consistent with the algebraic working.

Clearly, arg(z5) = 2/0 = (90 deg)

(Recall here that the value of tan(90) = oo!)

Some more problems for the next instalment on complex numbers, but first – another way to express polar forms.

A useful identity for application to complex numbers and the (cis) formula is:

cos(theta) + isin(theta) = r exp(i theta)

Thus, the previous numbers we divided (z1 and z2) may be expressed:

z1 = 1.4 [cos (45) + isin(45)] = 1.4 exp (i 45)

and

z2 = 2[cos(-30) +isin(-30)] = 2 exp(i(-30))

Using these results, do the following:

1.Express each of the following end results in the form r exp(itheta):

a) (2 + 3i)(1 – 2i)

b) (1 + i)/ (1- i)

c) (1 + [-3]^1/2)^2

Here, the resultant is first shown in the diagram, via graphical completion of the parallelogram formed by the vectors z3, and z4.. Readers can check to see it is consistent with the algebraic working.

Clearly, arg(z5) = 2/0 = (90 deg)

(Recall here that the value of tan(90) = oo!)

Some more problems for the next instalment on complex numbers, but first – another way to express polar forms.

A useful identity for application to complex numbers and the (cis) formula is:

cos(theta) + isin(theta) = r exp(i theta)

Thus, the previous numbers we divided (z1 and z2) may be expressed:

z1 = 1.4 [cos (45) + isin(45)] = 1.4 exp (i 45)

and

z2 = 2[cos(-30) +isin(-30)] = 2 exp(i(-30))

Using these results, do the following:

1.Express each of the following end results in the form r exp(itheta):

a) (2 + 3i)(1 – 2i)

b) (1 + i)/ (1- i)

c) (1 + [-3]^1/2)^2

2. Plot the results of (b) and (c) on the same Argand diagram and obtain the resultant. Check algebraically!

**Next: working with complex conjugates!**
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