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CA_19-35 - Tutorial CA A Workbook on Complex Arithmetic d...

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Tutorial CA: A Workbook on Complex Arithmetic d. j 3 +4 i j = 5; arg(3 +4 i ) = tan ¡ 1 (4 = 3) = 0 : 927 radians = ) 3+ 4 i = 5 e 0 : 927 i : e. j 3 ¡ 4 i j = 5; arg(3 ¡ 4 i ) = tan ¡ 1 ( ¡ 4 = 3) = 5 : 36 radians = ) 3 ¡ 4 i = 5 e 5 : 36 i : f. (1+ i ) = (1 ¡ i ) = 2 i= 2 = i: Thus, j (1 + i ) = (1 ¡ i ) j = 1; arg[(1+ i ) = (1 ¡ i )] = ¼= 2 = ) (1+ i ) = (1 ¡ i ) = e i¼= 2 : g. (1+ 2 i )(2 ¡ i ) = 4+ 3 i = 5 e : 644 i Exercise 19 From Eqn. 19, we can write cos μ + i sin μ = e ; cos μ ¡ i sin μ = e ¡ ; where the second equation is just the complex conjugate of the first (also obtained by μ ! ¡ μ: ) Solving, in turn, for cos μ and sin μ yields Eqns. 22 and 23. Exercise 20 a. 1 + i = p 2 e i¼= 4 ; 1 ¡ i = p 2 e 7 i¼= 4 = ) (1+ i )(1 ¡ i ) = p 2 £ p 2 £ e (1 = 4+7 = 4) = 2 e 2 ¼i = 2 since e 2 ¼i = 1 : The quotient is (1+ i ) (1 ¡ i ) = p 2 p 2 exp (1 = 4 ¡ 7 = 4) = e ¡ 3 i¼= 2 : b. 2 + 3 i = p 13 e 0 : 983 i ; 3 ¡ 2 i = p 13 e ¡ 0 : 588 i = ) (2 +3 i )(3 ¡ 2 i ) = p 13 £ p 13 £ e (0 : 983 ¡ 0 : 588) i = 13 e 0 : 395 i : The quotient is 2 +3 i 3 ¡ 2 i = p 13 p 13 exp i (0 : 983 ¡ ( ¡ 0 : 588)) = e 1 : 57 i : c. 5 = 5 e 0 i ; 3+ 4 i = 5 e 0 : 927 i = ) 5(3 +4 i ) = 25 e 0 : 927 i : The quotient is 5 5 exp i (0 ¡ 0 : 927) = e ¡ 0 : 927 i : Exercise 21 From Eqn. 19, we have e i ( μ 1 + μ 2 ) = cos( μ 1 + μ 2 )+ i sin( μ 1 + μ 2 ) = e 1 £ e 2 = (cos μ 1 + i sin μ 1 )(cos μ 2 + i sin μ 2 ) = (cos μ 1 cos μ 2 ¡ sin μ 1 sin μ 2 ) + i (cos μ 1 sin μ 2
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