Example 6

Express in the form a + bi

= (3 + i)

\   3  +  i   =   a  +  bi   where a  =  3 and b  =  1

b)

\        a = 1/5 and b =

Example 7

If   z₁   =   6  -  2i  and   z₂  =  2  - 5i    find   (a) | 2 z₁- 3z₂ |

Solution

2z₁ - 3z₂ = 2 (6 - 2i) - 3 ( 2 -5i) = (12 - 6) + (-4 + 15) i = 6 + 11i.

Now | 2z₁ - 3z₂ | =

(b) | z₁. z₂|

Solution

   z₁ . z₂   =   (6 - 2i) (2 - 5i) = 12 - 30i - 4i + 10i2

= 12 - 34i - 10 = 2 - 34i \ |z1 . z2 |
=

(c)

Solution

(d)

Solution

= 2.126 (approx.)

Example 8

Let a + bi =
so that a = -2 and b =
Now    

or = 120⁰

(b)     z =

Solution
a + bi =

(c)  If z1 = Find arg ( z1 . z2)

Solution

( z1 . z2) = = = -8 + 0i

\ arg ( z1 . z2) = tan^-1(0/-8) = 0 or p

Example 9

Express in the form a + bi.

(a)

Solution

Solution

 = a + bi

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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