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append - APPENDIX CN 1 Complex Numbers Complex Plane The...

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APPENDIX CN Complex Numbers ! Complex Plane 1. The complex numbers 3 3 + i , 4 i , 2, and 1 i are plotted as the respective points 3 3 , ( ) , 0 4 , ( ) , 2, 0 ( ) , and 1 1 , ( ) in the complex plane (see figure). –4 –4 4 4 Re( ) z Im( ) z (1, –1) (2, 0) (3, 3) (0, 4) ! Complex Operations 2. (a) 2 3 4 8 2 12 3 11 10 2 + ( ) ( ) = + = + i i i i i i (b) 2 3 1 2 2 3 3 1 5 2 + ( ) + ( ) = + + + = − + i i i i i i (c) Rationalizing the denominator, multiply the numerator and denominator by 1 i yielding 1 1 1 1 1 2 1 2 2 + = = i i i i i . (d) Rationalizing the denominator, we multiply the numerator and denominator by 3 i yielding 2 3 3 3 7 10 7 10 10 + + = + = + i i i i i i . ! Complex Exponential Numbers 3. (a) Using Euler’s formula, we write e i i i 2 2 2 1 0 1 π π π = + = + ( ) = cos sin . 793
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794 APPENDIX (b) Using Euler’s formula, we write e i i i i π π π 2 2 2 0 1 = + = + ( ) = cos sin . (c) Using Euler’s formula, we write e i i i = ( ) + ( ) = = − π π π π π cos sin cos sin 1. (d) Using the property e e e a b a b + = and using Euler’s formula, we write e e e e i e i e ie i i 2 4 2 4 2 2 2 2 4 4 2 2 2 2 2 2 2 2 + = = + F H I K = + F H G I K J = + π π π π a f cos sin . ! Magnitudes and Angles 4. (a) Absolute value: 1 2 1 2 5 2 2 + = + = i . Polar angle: θ = ° tan 1 2 1 63 or roughly 63 180 π radians. (b) Absolute value: = + − ( ) = i 0 1 1 2 2 . Polar angle: The complex number – i is located at the point 0 1 , ( ) in the complex plane so the angle is 3 2 π radians (or 270 ° ). (c) Absolute value: − − = ( ) + − ( ) = 1 1 1 2 2 2 i . Polar angle: θ π = + tan 1 1 b g and because the number − − 1 i is in the third quadrant in the complex plane, we have θ π = 5 4 radians (or 225 ° ). (d) Absolute value: + = ( ) + = 2 3 2 3 13 2 3 i . Polar angle: θ π = F H G I K J ° tan 1 3 2 124 or 124 180 π radians. (e) e i 2 . We write the exponential as e i i 2 2 2 = + cos sin .
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SECTION CN Complex Numbers 795 Magnitude is e i i 2 2 2 2 2 2 2 1 = + = + = cos sin cos sin . Polar angle is θ = F H I K = ( ) = tan sin cos tan tan 1 1 2 2 2 2. (f) 2 1 + + i i . We rationalize the denominator to get 2 1 1 1 3 2 2 + + = i i i i i . Magnitude is 2 1 3 2 1 2 1 2 10 2 2 + + = F H I K + F H I K = i i . Polar angle is θ = F H I K ≈ − ° tan . 1 1 3 184 or 341.6 ° . ! Complex Verification I 5. We check the first root z i = − + 1 by direct substitution: − + ( ) + − + ( ) + = + + = 1 2 1 2 1 2 1 2 2 2 0 2 i i i i . The second root − − 1 i is left to the reader. ! Complex Verification II 6. By direct substitution we have 1 2 1 4 1 1 4 1 1 1 4 1 2 1 1 2 1 1 4 4 1 4 4 2 2 2 + F H I K = + ( ) = + ( ) + ( ) = + ( ) + ( ) = = − i i i i i i i b g . ! Real and Complex Parts 7. Calling the complex number z a ib = + , we write z z a ib a ib a b iab a ib a b a i ab b 2 2 2 2 2 2 2 2 2 2 2 2 2 + = + ( ) + + ( ) = + + + ( ) = + + + ( ) b g b g . (a) Re z z a b a 2 2 2 2 2 + = + b g (b) Im z z b a 2 2 2 1 + = + ( ) b g
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796 APPENDIX ! Absolute Value Revisited 8. Using the formula z zz = , yields 4 2 4 2 4 2 16 4 2 5 + = + ( ) ( ) = + = i i i . ! Roots of Unity 9. The m roots of z m = 1 (called the roots of unity) are the m values z k m i k m k m = F H G I K J + F H G I K J F H G I K J 1 2 2 1 cos sin π π , k m = 01 1 , ! .
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