assignment5bfsol

# Assignment5bfsol - 1 ma2176 a5 Brief Sol Complex Numbers 1 Find the square roots of 5 12 i − Solution If 5 12 i a bi − = then 2 2 2 5 12 2 i a

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Unformatted text preview: 1 ma2176 a5 Brief Sol Complex Numbers 1. Find the square roots of 5 12 i − . Solution : If 5 12 i a bi − = + , then ( ) 2 2 2 5 12 2 i a bi a b abi − = + = − + . Equating the real and imaginary parts, we have 2 2 5 6 a b ab  − =  = −  . Thus, the solutions to 5 12 i − are 3 2 , 3 2 i i − − + . 2. Simplify ( ) 5 1 3 i − + . Solution : Write 2 3 1 3 2 i i e π − + = . Thus, ( ) ( ) 5 1 3 1 6 1 3 i i − + = − + . 3. Find the modulus and argument of ( ) ( ) 2 1 1 3 1 i i i + − + − + . Solution : The Euler forms of the complex numbers 1 , 1 3 and 1 i i i + − + − + are 2 3 3 4 4 2 , 2 and 2 i i i e e e π π π respectively. The modulus and argument of ( ) ( ) 2 1 1 3 1 i i i + − + − + are 2 2 and 5 12 π respectively. 4. If ω ( ) 1 ≠ is a complex cube root of unity, show that (a) ( )( ) 2 2 1 1 4 ω ω ω ω − + + − = . (b) ( ) ( ) 6 6 2 2 2 5 2 2 2 5 729 ω ω ω ω + + = + + = . Solution : ( )( ) 3 2 1 1 1 z z z z − = ⇔ − + + = . Suppose ( ) 1 ω ≠ is a complex cube root of unity, that is, ( )( ) 2 2 1 1 1 ω ω ω ω ω − + + = ⇒ + + = . (a) ( )( ) ( )( ) ( )( ) 2 2 2 2 2 2 3 1 1 1 2 1 2 2 2 4 4 ω ω ω ω ω ω ω ω ω ω ω ω ω − + + − = + + − + + − = − − = = . (b) ( ) ( ) ( ) ( ) ( ) 6 6 6 2 6 2 2 2 6 6 3 2 5 2 2 2 2 3 2 1 3 3 3 729 729 ω ω ω ω ω ω ω ω ω ω ω   + + = + + + = + + + = = = =   . ( ) ( ) ( ) ( ) 6 6 6 4 2 2 2 2 6 1 2 3 2 2 5 2 2 2 3 3 3 729 729 ω ω ω ω ω ω ω ω + + = + + + = = = = . 5. Find the modulus and argument of 1 cos sin 1 cos sin i i θ θ θ θ − − + + when 0 θ π < < ....
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## This note was uploaded on 11/09/2009 for the course ENG 91301 taught by Professor Lui during the Spring '08 term at Hong Kong Institute of Vocational Education.

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Assignment5bfsol - 1 ma2176 a5 Brief Sol Complex Numbers 1 Find the square roots of 5 12 i − Solution If 5 12 i a bi − = then 2 2 2 5 12 2 i a

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