6.3 Trigonometry and Complex Numbers



Let z = a + b i denote a complex number with real part a and imaginary part b. Then the absolute value of z, denoted | z |, is the distance from z to zero in the complex plane and is computed by the formula
absolute value of z equals square root of a square plus b square

The line containing 0 and z makes an angle t with the real axis satisfying

tan ( t ) = b / a. Furthermore,

cos ( t ) = a / | z | and

sin ( t ) = b / | z |. Thus

a = | z | cos ( t ) and b = | z | sin ( t ), so

z = | z | cos ( t ) + | z | sin ( t ) i . Thus

z = | z | [ cos ( t ) + i sin ( t ) ]

This is called the trigonometric form of a complex number.

Suppose z1 and z2 are complex numbers. Then their product is

z1 . z2 = | z1 | [ cos ( t1 ) + i sin ( t1 ) ] | z2 |[ cos ( t2 ) + i sin ( t2 ) ]

= | z1 | | z2 | {[ cos ( t1 ) cos ( t2 ) - sin ( t1 ) sin ( t2 ) ]
+ i [ cos ( t1 ) sin ( t2 ) + sin ( t1 ) cos ( t2 ) ] }

= | z1 | | z2 | [ cos ( t1 + t2 ) + i sin ( t1 + t2 ) ]. Thus,

z1 . z2 = | z1 | | z2 | [ cos ( t1 + t2 ) + i sin ( t1 + t2 ) ]

Exercise 6.3.1

Let z = 3 + 3 i. Find | z |. Find an angle t such that z = | z | [ cos ( t ) + i sin ( t ) ]. Write z in trigonometric form.

Solution

Exercise 6.3.2

Show that z2 = | z |2 [ cos ( 2 t ) + i sin ( 2 t ) ].

Solution

The result in Exercise 6.3.2 can be extended to the general result, called DeMoivre's Theorem:

If z = | z | [ cos ( t ) + i sin ( t ) ], then zn = | z |n [ cos ( n t ) + i sin ( n t ) ].

Example: Find ( -1 + 2 i )5

Let z = -1 + 2 i. Then find z5. First, we must find the absolute value of z is the square root of 5which is approximately 2.236067977. Next, we must find the angle t. Since z lies in the second quadrant of the complex plane, we are looking for an angle t in the second quadrant satisfying tan ( t ) = b / a = -2. arctan ( -2 ) will not do, since it is in quadrant IV. We want pi + arctan ( -2 ), which is approximately 2.034443936. [Your calculator must be in radian mode.] Thus, ( -1 + 2 i )5 is approximately equal to

( 2.236067977 )5 [ cos 10.17221968 + i sin 10.17221968 ]

= 55.90169937 [ -0.733430296 - 0.679764665 i ]

= - 40.99999992 - 37.99999999 i.

Since the real and imaginary parts of ( -1 + 2 i )5 must be integers, the answer is

( -1 + 2 i )5 = - 41 - 38 i.

Exercise 6.3.3

Find ( 1 + i )7.

Solution

The inverse of finding powers of complex numbers is finding roots of complex numbers.

A complex number has two square roots, three cube roots, four fourth roots, etc. Generally, a complex number has n nth roots. All n of the nth roots of a complex number z are evenly spaced around a circle centered at 0 and having a radius equal to the nth root of the absolute value of z. Furthermore, if z = | z | [ cos ( t ) + i sin ( t ) ], then one of the n nth roots of z is | z |1 / n [ cos ( t / n ) + i sin ( t / n ) ]. Since the n nth roots are evenly spaced around a circle, they are separated by an angle of two pi over n. So, once we find the root | z |1 / n [ cos ( t / n ) + i sin ( t / n ) ], we find the other n - 1 roots by adding multiples of two pi over n to t / n .

Example: Find the cube roots of 1 + i.

First, write 1 + i in trigonometric form: square root of two times sum of cosine pi over 4 and i times sine pi over 4

Then one root is sixth root of 2 times sum of cosine pi over 12 and i times sine pi over 12. To find the two remaining cube roots, we add multiples of two pi over 3 to pi over 12 to get three pi over 4 and 5 pi over 6. Thus the three cube roots of 1 + i are

sixth root of two times sum of cosine pi over 12 and i times sine pi over 12

sixth root of 2 times sum of cosine 3 pi over 4 and i times sine of 3 pi over 4

sixth root of 2 times sum of cosine 5 pi over 6 and i times sine of 5 pi over 6

Exercise 6.3.4

Find the four fourth roots of - 16 i.

Solution

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