Camera Maintenance & Repair, Book 2: Fundamental Techniques: by Thomas Tomosy

By Thomas Tomosy

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The geometrical effect of this in an Argand diagram can be seen using the Flash resource Multiplication and division in the Argand diagram. This topic is covered more fully in FP2. © MEI, 19/07/10 7/7 Further Pure Mathematics 1 Trigonometry needed for Complex numbers In your work on complex numbers, you need to be able to find the argument of a complex number, which may involve dealing with angles greater than 90°. If you haven’t yet done the work on Trigonometry in C2, or the chapter on Matrices in FP1, this may be new to you.

10. You will also meet them in C4. w r1 (cos 1  jsin 1 )  z r2 (cos  2  jsin  2 )  r1 (cos 1  jsin 1 )(cos  2  jsin  2 ) r2 (cos  2  jsin  2 )(cos  2  jsin  2 )  r1 (cos 1 cos  2  jcos 1 sin  2  jsin 1 cos  2  sin 1 sin  2 ) (cos 2  2  sin 2  2 ) cos²θ + sin²θ = 1 r  1   cos 1 cos  2  sin 1 sin  2   j  sin 1 cos  2  cos 1 sin  2   r2  r1  cos 1  2   jsin 1  2   r2 © MEI, 19/07/10 6/7 MEI FP1 Complex nos section 3 Notes and Examples wz  r1r2 , w r1  w  , arg( wz )  1   2 , arg    1   2 .

1 + j)2 = 1 + 2 j − 1 = 2 j (1 + j)3 = 2 j(1 + j) = 2 j − 2 = −2 + 2 j Substituting into z 3 − 2 z + k = 0 : −2 + 2 j − 2(1 + j) + k = 0 −2 + 2 j − 2 − 2 j + k = 0 k =4 1 + j is a root, so 1 – j is also a root. So a quadratic factor is ( z − 1 − j)( z − 1 + j) = ( z − 1)2 + 1 z 3 − 2 z + 4 = ( z 2 − 2 z + 2)( z + 2) = z 2 − 2z + 2 so the other two roots are 1 – j and -2. © MEI, 21/12/06 1/3 Further Pure Mathematics 1 4. z = −1 + j z 2 = ( −1 + j)2 = 1 − 2 j − 1 = −2 j z 3 = −2 j( −1 + j) = 2 j + 2 = 2 + 2 j z 4 = (2 + 2 j)( −1 + j) = −2 − 2 = −4 Substituting into z 4 − 2 z 3 − z 2 + 2 z + 10 : −4 − 2(2 + 2 j) − ( −2 j) + 2( −1 + j) + 10 = −4 − 4 − 4 j + 2 j − 2 + 2 j + 10 =0 so -1 + j is a root.

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