1. CARTESIAN COMPLEX NUMBERS - Weebly

UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 1. CARTESIAN COMPLEX NUMBERS introduction Try to solve this quadratic equation : 0522=++xx By using quadratic formula : the discriminant , 16)5)(1(4)2(422 = = = acb the solution : )1(216)2( =x but it is not possible to evaluate 1 however if an operator j is defined as then the solution may be expressed as : 12 =j 21242)1(216)2(jjx = = = 21j+ and are known as COMPLEX NUMBERS . 21j Both solutions are of the form : = x1 2j COMPLEX number Real part Imaginarypart z = ajb this form is known as the CARTESIAN COMPLEX NUMBERS ( ALGEBRAIC FORM ) E2 - 1 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 EXAMPLES EXAMPLE 1 : Solve the quadratic equation , 042=+x 240422jxxx = ==+ EXAMPLE 2 : Solve the quadratic equation , 2 x 2 + 3 x + 5 = 0 2343463436344093)2(2)5)(2(4332jxjxxxx = = = = = E2 - 2 - MATHEMATICS UNIT UNIVERSITI KUALA LUMPUR COMPLEX NUMBER E2 POWERS OF j j 0 0)1( 1 j 1 1 j j 2 )1)(1( -1 j 3 j2j = (-1)j -j j 4 j2j2 = (-1)(-1) 1 j 5 j4j = (1)j j In genera

4.1 INTRODUCTION Let a complex number Z = a + jb as shown in the Argand Diagram below. Let the distance OZ be r and the angle OZ makes with the positive real axis be θ. Imaginary axis Z r jb θ Real axis O a A From the trigonometry of right angle triangle :

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  Introduction, Cartesian

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