complex numbers assignment 5 solutions

complex numbers assignment 5 solutions - Assignment 5:...

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2. (1 – i) z = 1 – 3i z = i 1 i 3 1 - (M1) z = i 1 i 1 i 1 i 3 1 + + × - (M1) z = 2 – i (A2) OR Let z = x + i y (1 – i)( x + i y ) = 1 – 3i (M1) x + y – i( x y ) = 1 – 3i = - = + 3 1 y x y x (M1) x = 2, y = –1 (A2)(C2)(C2) Note: Award (C4) for z = 2 i. [4] 4. (a) z 1 = 2 2 i 6 - | z 1 | = 4 2 4 6 + (M1) = 2 (A1) arg z 1 = arctan 6 π 3 1 - =  - (A1) Therefore, z 1 = - + - 6 π sin i 6 π cos 2 (C1) z 2 = 1 – i | z 2 | = 2 1 1 = + arg z 2 = arctan (–1) = – 4 π (A1) z 2 = + - 4 π sin i 4 π cos 2 (C1) 6 (b) + - + - = 4 π sin i 6 π cos 2 6 π sin i 6 cos 2 2 1 π z z = 1 + + + - 4 π 6 π sin i 4 π 6 π cos (M2) = cos 12 π + i sin 12 π (AG) 2 1
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complex numbers assignment 5 solutions - Assignment 5:...

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