*Solution to last problem - obtaining the resultant for answers (b) and (c).*We will get into some new areas with the algebra of complex numbers in this instalment. But first things – let’s do the problems from the end of the last one.

We want to express each of the following in the polar-exponential form: r exp(i theta):

a) (2 + 3i)(1 – 2i)

Multiplying we get: 2 + 3i -6(i^2) -4(i) = 8 + 3i – 4i = 8 –i

First, convert to standard polar form:

z = x + iy

so: r = [8^2 + (-1)^2]^1/2 = 8.1

theta = arg(z) = arctan (y/x) = arctan (-1/8) = -7.1 deg.

So: z = 8.1 [cos(-7.1) + isin(-7.1)]

Or: z = 8.1 exp (i×(-7.1))

b) (1 + i)/ (1 – i)

We already have:

(1 + i) = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)

Working done in text of Part (IV).

So, z1 = 1.4 exp (i×45)

The reader should be able to easily show that (1- i) is the same except for the argument, which must be (- theta) since we have arctan (-1/1).

Thus:

z2 = (1 – i) = 1.4 exp (i×(-45))

Then: dividing:

(z1/z2) = (1.4 cis(45)/ 1.4 cis(-45)) = 1.4 cis{90}= 1.4 [cos(90) + isin(90)] = 1.4 isin(90) = 1.4i

Or z = 1.4 exp (i×90)

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

expand the quantity on the right side:

z = [1 + (-3)^1/2] x [1 + (-3)^1/2]

But bear in mind that: [(-3)^1/2] = [i(3)^1/2] so we have:

(1 + [i(3)^1/2]) x (1 + [i(3)^1/2]) = 1 + 2i[(3)^1/2] + (-1)3 = -2 + i2[(3)^1/2] = -2 + 3.46i

Whence: r = [(-2)^2 + (2[(3)^1/2)^2] = [4 + 36]^1/2 = [40]^1/2 = 6.3

And theta = arctan (y/x) = arctan (2[(3)^1/2]/ -2) = arctan (-(3)^1/2]) = arctan (-1.732) = (-60)

Thus, z = 6.3 exp(i×(-60))

We plot the results of (b) and (c) on the Argand diagram and obtain the resultant

Call z(b) = 0 + 1.4i (from that solution)

And z(c) = -2 + 3.46i

The plots of the two numbers are shown in the diagram. The resultant is sketched and can be confirmed from the algebra:

z(b) + z(c)= 1.4i + (-2 +3.46i) = -2 + 4.86i

Now, we go to complex conjugates – extremely important – especially in applications to the quantum wave functions for which probability amplitudes are obtained using the complex conjugate of the psi- wave.

To obtain the complex conjugate is very simple- for a complex number of the form

z = x + iy

The complex conjugate is written: z* = x – iy

Let’s use this to obtain the complex conjugates of the complex numbers in problems 1(a)-(c).

Thus for (a): (2 + 3i)(1 – 2i) = (8 –i)

But multiplying their complex conjugates together would mean

(2 - 3i)(1 + 2i)

Would one obtain the complex conjugate of the original result, e.g. (8 + i) – with no need to work it out?

Check!

(2 - 3i)(1 + 2i) = 2 + i -6(i^2) = 8 + i

It works.

b) The complex conjugate for this problem would be:

(1 + i)/ (1 – i) = (1 – i)/(1 + i)

What result would you get? How different?

Lastly, for part(c) the result we obtained in x + iy form was -2 + 3.46i

The complex conjugate is:

-2 – 3.46i

Note the first member (x-value) never changes!

A couple more odds and ends:

For a complex number z = x + iy

The REAL part of z denoted by Re(z) is always x!

The imaginary part of z, denoted Im(z) is the real number y! (Not iy!)

We’ll use these points and more when next we return to the fascinating realm of complex numbers.

We want to express each of the following in the polar-exponential form: r exp(i theta):

a) (2 + 3i)(1 – 2i)

Multiplying we get: 2 + 3i -6(i^2) -4(i) = 8 + 3i – 4i = 8 –i

First, convert to standard polar form:

z = x + iy

so: r = [8^2 + (-1)^2]^1/2 = 8.1

theta = arg(z) = arctan (y/x) = arctan (-1/8) = -7.1 deg.

So: z = 8.1 [cos(-7.1) + isin(-7.1)]

Or: z = 8.1 exp (i×(-7.1))

b) (1 + i)/ (1 – i)

We already have:

(1 + i) = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)

Working done in text of Part (IV).

So, z1 = 1.4 exp (i×45)

The reader should be able to easily show that (1- i) is the same except for the argument, which must be (- theta) since we have arctan (-1/1).

Thus:

z2 = (1 – i) = 1.4 exp (i×(-45))

Then: dividing:

(z1/z2) = (1.4 cis(45)/ 1.4 cis(-45)) = 1.4 cis{90}= 1.4 [cos(90) + isin(90)] = 1.4 isin(90) = 1.4i

Or z = 1.4 exp (i×90)

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

expand the quantity on the right side:

z = [1 + (-3)^1/2] x [1 + (-3)^1/2]

But bear in mind that: [(-3)^1/2] = [i(3)^1/2] so we have:

(1 + [i(3)^1/2]) x (1 + [i(3)^1/2]) = 1 + 2i[(3)^1/2] + (-1)3 = -2 + i2[(3)^1/2] = -2 + 3.46i

Whence: r = [(-2)^2 + (2[(3)^1/2)^2] = [4 + 36]^1/2 = [40]^1/2 = 6.3

And theta = arctan (y/x) = arctan (2[(3)^1/2]/ -2) = arctan (-(3)^1/2]) = arctan (-1.732) = (-60)

Thus, z = 6.3 exp(i×(-60))

We plot the results of (b) and (c) on the Argand diagram and obtain the resultant

Call z(b) = 0 + 1.4i (from that solution)

And z(c) = -2 + 3.46i

The plots of the two numbers are shown in the diagram. The resultant is sketched and can be confirmed from the algebra:

z(b) + z(c)= 1.4i + (-2 +3.46i) = -2 + 4.86i

Now, we go to complex conjugates – extremely important – especially in applications to the quantum wave functions for which probability amplitudes are obtained using the complex conjugate of the psi- wave.

To obtain the complex conjugate is very simple- for a complex number of the form

z = x + iy

The complex conjugate is written: z* = x – iy

Let’s use this to obtain the complex conjugates of the complex numbers in problems 1(a)-(c).

Thus for (a): (2 + 3i)(1 – 2i) = (8 –i)

But multiplying their complex conjugates together would mean

(2 - 3i)(1 + 2i)

Would one obtain the complex conjugate of the original result, e.g. (8 + i) – with no need to work it out?

Check!

(2 - 3i)(1 + 2i) = 2 + i -6(i^2) = 8 + i

It works.

b) The complex conjugate for this problem would be:

(1 + i)/ (1 – i) = (1 – i)/(1 + i)

What result would you get? How different?

Lastly, for part(c) the result we obtained in x + iy form was -2 + 3.46i

The complex conjugate is:

-2 – 3.46i

Note the first member (x-value) never changes!

A couple more odds and ends:

For a complex number z = x + iy

The REAL part of z denoted by Re(z) is always x!

The imaginary part of z, denoted Im(z) is the real number y! (Not iy!)

We’ll use these points and more when next we return to the fascinating realm of complex numbers.

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