PureMath.pdf

# First to prove that z z 5 z z is to prove that x x 2

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[First, to prove that | z + z 0 | 5 | z | + | z 0 | is to prove that ( x + x 0 ) 2 + ( y + y 0 ) 2 5 { p x 2 + y 2 + p x 0 2 + y 0 2 } 2 . The theorem is then easily extended to the general case.] 2. The one and only case in which | z | + | z 0 | + · · · = | z + z 0 + . . . | , is that in which the numbers z , z 0 , . . . have all the same amplitude. Prove this both geometrically and analytically. 3. The modulus of the sum of any number of complex numbers is not less than the sum of their real (or imaginary) parts. 4. If the sum and product of two complex numbers are both real, then the two numbers must either be real or conjugate. * The numbers a + b , a - b , where a , b are rational, are sometimes said to be ‘conjugate’.

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[III : 46] COMPLEX NUMBERS 106 5. If a + b 2 + ( c + d 2) i = A + B 2 + ( C + D 2) i, where a , b , c , d , A , B , C , D are real rational numbers, then a = A, b = B, c = C, d = D. 6. Express the following numbers in the form A + Bi , where A and B are real numbers: (1 + i ) 2 , 1 + i 1 - i 2 , 1 - i 1 + i 2 , λ + μi λ - μi , λ + μi λ - μi 2 - λ - μi λ + μi 2 , where λ and μ are real numbers. 7. Express the following functions of z = x + yi in the form X + Y i , where X and Y are real functions of x and y : z 2 , z 3 , z n , 1 /z , z +(1 /z ), ( α + βz ) / ( γ + δz ), where α , β , γ , δ are real numbers. 8. Find the moduli of the numbers and functions in the two preceding examples. 9. The two lines joining the points z = a , z = b and z = c , z = d will be perpendicular if am a - b c - d = ± 1 2 π, i.e. if ( a - b ) / ( c - d ) is purely imaginary. What is the condition that the lines should be parallel? 10. The three angular points of a triangle are given by z = α , z = β , z = γ , where α , β , γ are complex numbers. Establish the following propositions: (i) the centre of gravity is given by z = 1 3 ( α + β + γ ); (ii) the circum-centre is given by | z - α | = | z - β | = | z - γ | ; (iii) the three perpendiculars from the angular points on the opposite sides meet in a point given by R z - α β - γ = R z - β γ - α = R z - γ α - β = 0; (iv) there is a point P inside the triangle such that CBP = ACP = BAP = ω,
[III : 46] COMPLEX NUMBERS 107 and cot ω = cot A + cot B + cot C. [To prove (iii) we observe that if A , B , C are the vertices, and P any point z , then the condition that AP should be perpendicular to BC is (Ex. 9) that ( z - α ) / ( β - γ ) should be purely imaginary, or that R ( z - α ) R ( β - γ ) + I ( z - α ) I ( β - γ ) = 0 . This equation, and the two similar equations obtained by permuting α , β , γ cyclically, are satisfied by the same value of z , as appears from the fact that the sum of the three left-hand sides is zero. To prove (iv), take BC parallel to the positive direction of the axis of x . Then * γ - β = a, α - γ = - b Cis( - C ) , β - α = - c Cis B. We have to determine z and ω from the equations ( z - α )( β 0 - α 0 ) ( z 0 - α 0 )( β - α ) = ( z - β )( γ 0 - β 0 ) ( z 0 - β 0 )( γ - β ) = ( z - γ )( α 0 - γ 0 ) ( z 0 - γ 0 )( α - γ ) = Cis 2 ω, where z 0 , α 0 , β 0 , γ 0 denote the conjugates of z , α , β , γ . Adding the numerators and denominators of the three equal fractions, and using the equation i cot ω = (1 + Cis 2 ω ) / (1 - Cis 2 ω ) , we find that i cot ω = ( β - γ )( β 0 - γ 0 ) + ( γ - α )( γ 0 - α 0 ) + ( α - β )( α 0 - β 0 ) βγ 0 - β 0 γ + γα 0 - γ 0 α + αβ 0 - α 0 β .

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