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Example 7

Find the continued product of all values of

Solution

1 + = cos (2 p k + p / 3) + sin (2 p k + p / 3)

Note that r = = 1 and q = tan-1(Ö 3 / 2) = 600 or p/3c

putting k = 0, 1, 2, 3, we get

Therefore the continued product is

\ z1 . z2 . z3 . z4 = cos (10p) + i sin (10p) = cos (0) + i sin (0) = 1 + i (0) = 1

Example 8

If l + 2i is one root of x4 - 3x3 + 8x2 - 7x +5 = 0. Find the other roots.

Solution

Let a = = 1 + 2i then b = 1 -2i

We know that a quadratic equation in x, having roots a and b, can be put in the form

x2 - ( a + b ) x + a . b = 0

\ x2 - [ (1 + 2i) + (1 - 2i) ] x + (1 + 2i) (1 - 2i) = 0

\ x2 - 2x + 5 = 0

Dividing by x2 - 2x + 5 = 0 to x4 - 3x3 + 8x2 - 7x + 5 = 0 as

Since R = 0, x2 - x + 1 = 0 is the other factor of the given expression.

Thus the other factor is x2 - x + 1 whose roots are

Now x =

Therefore x4 - 3x3 + 8x2 - 7x + 5 = 0 has roots

1 + 2i, 1 - 2i,

Example 9

Find two values of

Solution

i = cos 900 + i sin 900....[ since cos 900 = 0 and sin 900 = 1]

Example 10 Find the two values of the sq.root of

Solution

1 + 2i Þ a = 1 > 0 and b = 2 > 0 i.e. (q is in 1st quad.)

\ r =

= 63026'

Also, 2 + i Þ a = 2 > 0 and b = 1 > 0 i.e. (q is in 1st quad.)

\ r = Ö4+1 = Ö5 and q = tan-1(1/2) = 26034'

\ = cos (63024' - 26036') + i sin (63024' - 26036' )

      = cos (36052') + i sin (36052')

Now

= cos 36052' + 2pk + i sin 2pk + 36052'

....... by using z1/n = r1/n

When k = 0

\ = cos (18026') + i sin (18026') = 0.9487 + 0.3162i

When k = 1

= cos (198026') + i sin (198026')

= - 0.9487 - 0.3162i

Thus required sq.roots are ± (0.9487 + 0.3162i)

Example 11 The centre of a regular hexagon is at the origin and one vertex is given by (1+i) on the Argand's diagram. Find the remaining vertices.

Solution

Since represent Aº (1+ i) the first vertex of the regular hexagon ABCDEF.
Now 1+ i Þ a=1 and b=1 (i.e. q in 1st quad.). We have r= = units i.e. || = units and q = tan-1(1/1) = p/4 = 450. \ makes 450 angle with and of length units. Being regular hexagon , , , , q make angles of p/3, 2p/3, p, 4p/3, 5p/3 with . This shows that modulus of all these six vertices is units and their respective amplitudes are, (p/4+ p/3), (p/4+2p/3), (p/4+p), (p/4+5p/3) for B, C, D, E and F.

\ Radius vectors representing the vertices are

A,

B,

C,

D,

E, and

F,

i.e. A is (1+i), B is (-0.366 + 1.366i), C is (-1.366 + 0.366i), D is (-1-i), E is (0.366 - 1.366i) and F is (1.366 - 0.366i) .

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Index

8.1 Geometry of complex numbers
8.2 De - Moivres's theorem
8.3 Roots of complex numbers
8.4 Cirsular functions of complex angles & hyperbolic function
Supplementary Problems

Chapter 9

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