By C. W. Celia, A. T. F. Nice, K. F. Elliott

ISBN-10: 0333348273

ISBN-13: 9780333348277

ISBN-10: 1349067113

ISBN-13: 9781349067114

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**Example text**

Then Z2 - ZI = 4 + 4i, and the equation of the line is Z = 2 - i + /(4 + 4i). The perpendicular bisector of a line In Fig. 8, the points P, A and B represent the complex numbers z, ZI and Z2 respectively. M is the mid-point of AB and PM is at right angles to AB. For any point P on the perpendicular bisector of AB, the lengths of AP and BP are equal. Since AP = [z - zll and BP = [z - z21, this gives [z - zii = [z - z21· This is the equation of the line PM . Complex numbers 39 y A B x Example 2 Write down the equation of the perpendicular bisector of the line joining the points representing 2 - 3i and 4 + i.

Let Z = (I + i)z and w = Z + 2 + i. Since 1 + i = J 2(cos n/4 + i sin n/4), the transformation from the z-plane to the Z-plane gives a magnification by a scale factor ) 2, centre 0, followed by an anticlockwise rotation about the origin through n/4 radians. The transformation from the Z-plane to the w-plane is a translation which moves the origin to the point 2 + i. This is illustrated in Fig. 18. 18 BI 0, Z-plane u X w-plane Z= (1 + i) z. w= Z+ 2 + i Example 2 Express the transformation w = 3iz - 6 as the combination of a magnification, a rotation and translation in that order.

Method 1 = = Let 21 + 20i = R(cos a + i sin a) R2 = 2 F + 202 = 841 R = 29. The two square roots will be ± j29[cos (ta) By equating real parts, + i sin (ta)]. r A p 1 Q Fig. 1 26 Advanced mathematics 3 Square roots x 29 cos a = 21 cos a = 21/29 . = By equating imaginary parts, 29 sin a = 20 sin a = 20/29. = Since cos a and sin a are both positive, 0 < a < lr/2 and cos be both positive. cos a = 2 = = cos- ta - 1 = 21/29 cos- fa = 25/29 cos ta = 5/J29 sin ta = 2/J29. = Hence the two square roots are ± (5 Method 2 ta and sin ta will + 2i).

### Advanced Mathematics 3 by C. W. Celia, A. T. F. Nice, K. F. Elliott

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