EXERCISES

1. Plot the complex number 2 = 3 - 4j. Find its amplitude and modulus and express z in the polar form.

Solution. Plot the point B with coördinates (3, −4) as in Fig. 111. Then OB 3 4j and r = √x2 + y2

94. Multiplication of complex numbers when expressed in the polar form. Let 21 = ricos 01+j sin 01],

But this is the polar form of a complex number in which the modulus is r12 and the amplitude 01 + 02.

By the same reasoning, the product of

and

2122rr2 [cos (01 +02) + j sin (01 +02)],

23 = r3 (COS 03 +j sin 03),

gives 212223 = rir2rs [Cos (01+02+03) +j sin (01 + 02 + 03)],

and so on for any number of factors.

Hence the product of any number of complex numbers is a complex number which has for its modulus the product of the moduli, and for its amplitude the sum of the amplitudes of the factors.

(3)

That is, for n a positive integer,

[r (cos +j sin 0)]" = pn [cos no +j sin n☺].

This is known as DeMoivre's Theorem, for n a positive integer. Now suppose n a negative integerm, then

The following exercises are to be solved by means of DeMoivre's Theorem.

1. Show that sin 2 0 2 sin 0 cos 0 and cos 20

Solution. In (3) let n = 2 and r = 1.

Then

=

cos2 - sin2 0.

[cos 0+j sin 0]2 = cos 20+j sin 2 0.

Expanding the first member,

cos20 + 2j sin 0 cos 0 + j2 sin2 0 cos 20+j sin 2 0.

-1, therefore (cos2 0 sin2 ) + 2 sin ◊ cos ◊ • j

= cos 20+j sin 2 0.

By Art. 91, the real parts are equal and the coefficients of j are equal. That is, cos 20 cos2 0 — sin2 0 and sin 20 2 sin 0 cos 0.

(cos 45° + j sin 45°)2 = cos 90° + j sin 90° = j, since cos 90° = 0 and sin 90° = 1.

r1 cos 01 cos 02 + sin 01 sin 02 + j (sin 01 cos 02

cos 01 sin (2)

Hence the quotient of two complex numbers is a complex number having an amplitude equal to the amplitude of the dividend minus the amplitude of the divisor, and a modulus equal to the modulus of the dividend divided by the modulus of the divisor.

For example,

8 (cos 60°+j sin 60°)

4 (cos 30° + j sin 30°)

2 [cos (60° — 30°) + j sin (60° — 30°)]

2 (cos 30° + j sin 30°).

96. Square roots of a number. Let 22 = r2 (cos + j sin 0) be the number. Now the square root of a number is one of the two equal factors of the number. By Art. 94, when a number is squared the modulus is squared and the amplitude is doubled.

Therefore, in extracting the square root, we extract the positive square root of the modulus and divide the amplitude by 2.

Now cos 0 = cos (0 + 2 kπ) and sin 0

=

sin (0 + 2 kπ). Then the most general form in which the original number may be written

These are the only square roots, for if k = 2, the amplitude

becomes + 2 π, the functions of which are the same as those of

Ө 2

Ꮎ 2

If k 3, we have + 3, the functions of which are the same as

Ꮎ

2

+π, etc. Any other values of k will give either 20 or 21.

Example. Extract the square root of the number 1.

Solution. The complex number 1 is plotted on the x-axis, one unit to the right of the origin. The modulus is 1 and the amplitude is 0 + 2km.

97. Cube roots of a number. The cube root of a number is one of the three equal factors of the number. The number is the cube of this factor. Since in raising a number to the third power, we cube the modulus, and multiply the amplitude by 3, then in extracting the cube root, we extract the cube root of the modulus and divide the amplitude by 3.

Suppose 23 = r (cos + j sin ) is the number. Writing the number in the most general form we have

23 = r [cos (0 + 2 kπ) + j sin (0 + 2 kπ)].

roots corresponding to k = 0 and k = 3 are the same. Any other values of k will give an amplitude with the functions equal to those Therefore there are but three cube roots of They are

for k = 0, 1, or 2.

r [cos +j sin 0].