Example 2 Solve the equation z 2 - 2 z + 2 = 0

• 11
• 100% (2) 2 out of 2 people found this document helpful

This preview shows page 1 - 5 out of 11 pages.

Complex Analysis Chapter 1. Complex Numbers § 1.1. Algebraic Properties Definition. A complex number z is a pair of real numbers: z = ( x, y ) where x is the real part and y is the imaginary part of z : Re z = x, Im z = y. real number: x = ( x, 0), pure imaginary number: (0 , y ) z = 0 ⇐⇒ x = 0 y = 0 ⇐⇒ x 2 + y 2 = 0 The sum z 1 + z 2 and the product z 1 z 2 of the two complex numbers z 1 = ( x 1 , y 1 ) and z 2 = ( x 2 , y 2 ) are defined as follows ( x 1 , y 1 ) + ( x 2 , y 2 ) = ( x 1 + x 2 , y 1 + y 2 ) , ( x 1 , y 1 )( x 2 , y 2 ) = ( x 1 x 2 - y 1 y 2 , y 1 x 2 + x 1 y 2 ) . Denote i = (0 , 1). Then z = x + iy . i 2 = (0 , 1)(0 , 1) = ( - 1 , 0) , or i 2 = - 1 . The properties of addition and multiplication of complex numbers are the same as for real numbers: z 1 + z 2 = ( x 1 + iy 1 ) + ( x 2 + iy 2 ) = x 1 + x 2 + ( y 1 + y 2 ) i, z 1 z 2 = ( x 1 + iy 1 )( x 2 + iy 2 ) = x 1 x 2 - y 1 y 2 + ( x 1 y 2 + x 2 y 1 ) i. 1
Example 1. z 1 z 2 = 0 ⇐⇒ z 1 = 0 or z 2 = 0. Proof. z 1 z 2 = 0 ⇐⇒ x 1 x 2 - y 1 y 2 = 0 x 1 y 2 + x 2 y 1 = 0 ⇐⇒ ( x 1 x 2 - y 1 y 2 ) 2 + ( x 1 y 2 + x 2 y 1 ) 2 = 0 ⇐⇒ ( x 2 1 + y 2 1 )( x 2 2 + y 2 2 ) = 0 ⇐⇒ x 2 1 + y 2 1 = 0 or x 2 2 + y 2 2 = 0 ⇐⇒ z 1 = 0 or z 2 = 0 . Example 2. Solve the equation z 2 - 2 z + 2 = 0.
2
Thus, z - 1 = x x 2 + y 2 , - y x 2 + y 2 = x x 2 + y 2 - y x 2 + y 2 i. Division. z 1 z 2 = z 1 · z - 1 2 when z 2 6 = 0. Example 3. 3 - 4 i 1 + i = (3 - 4 i ) · 1 2 - 1 2 i · = - 1 2 - 7 2 i Example 4. 10 (1 + i )(2 + i )(3 + i ) = 10 (1 + 3 i )(3 + i ) = 10 10 i = 1 i = - i § 1.2. Geometric Properties Complex plane : z -plane C 1 i x y z=(x, y) = x + iy Vector in R 2 : ( x, y ) = x + iy = z C x y [ 3 [ 2! , ! [ 3 ; 2 x y [ 3 ; 2 [ 2 .[ 3 3
Modulus (absolute value) of z : | z | = p x 2 + y 2 distance to the origin Then | z | 2 = x 2 + y 2 . | z | = 0 ⇐⇒ x = y = 0 ⇐⇒ z = 0 , | z 1