10. Prove that the sum of the complex numbers representing the sides of a polygon taken in order equals zero. II. Three forces of 151, 106, and 61 horse power respectively, make angles of 50° 04' 30", 211° 20′ 305", and - 96° respectively with the x-axis Show that the resultant is zero, that is, that the forces are in equilibrium. 150. Multiplication and Division of Complex Numbers. Let Z1 r1 (cos 01 + i sin 01), Z2 = 12 (COS 02 + i sin 02) represent any two complex numbers. Their product is 2122712 (cos 01 + i sin 01) (cos 02 + i sin 02) = = rir2 [(cos 01 Cos 02 - sin 01 sin 02) + i (sin 01 cos 02+ cos 01 sin 02)] r12 [cos (01 + 02) + i sin (01 + 02)]. 1 1 r2 cos 02+ i sin 02 02 01 1 71 (cos 01 cos 01⁄2 + sin 01 sin 02) + i (sin 0, cos 02 — cos 01 sin 02), = 12 = 12 1 (cos2012 sin20) = 1 [cos (01-02) + i sin (01-02)]. From equations (1) and (2) it appears that, I (1) (2) The product of two complex numbers is another complex number whose modulus is the product of the moduli and whose argument is the sum of the arguments of the numbers. The quotient of two complex numbers is another complex number whose modulus is the modulus of the dividend divided by the modulus of the divisor, and whose argument is the argument of the dividend diminished by the argument of the divisor. Corollary 1. Since I = cos o° + i sin o°, i = cos 90° + i sin 90°, - I = cos 180° + i sin 180°, the modulus in each case being unity, it follows that: multiplying any complex number by I leaves it unchanged, multiplying any complex number by i increases its argument by 90°, multiplying any complex number by I increases its argument by 180°. In the first case the line segment representing the complex number is left unchanged, in the second case the line segment is turned in the positive direction through an angle of 90°, in the third case the line segment is reversed. Corollary 2. (cos + i sin 0) (cos i sin 0) 0 = 1; therefore cos + i sin and cos i sin @ are reciprocals, and in general the I reciprocal of r (cos ℗ + i sin 0) is (cos 0 — i sin 0). r + i sin 0) and r (cos 0 i sin 0) are Two numbers such as r (cos said to be conjugate to each other. the other. Each is called the conjugate of 151. Powers of Complex Numbers. By (1), Article 150, we have 2122 r1 (cos 01 + i sin 01) X 12 (cos 02 + i sin 02) = r1r2 [cos (01 +02) + i sin (01+ 02)]. Let us multiply this result by a third complex number z3, thus: = Z12223 rire [cos (01 +02) + i sin (01 +02)] X 73 (cos 03 + i sin 03) =r1r2r3 [COS (01 + 02 +03) + i sin (01 + 02 +03)]. Similarly we obtain for the product of n factors 21=r1 (cos 01 + i sin 01), z2 = 12 (cos 02+ i sin 02), The modulus of the product of any number of complex numbers is equal to the product of the moduli of the factors, and the argument of the product is equal to the sum of the arguments of the factors. Now let us suppose that the n factors in (1) are all equal, each factor being Equation (3) embodies one of the most famous theorems of modern analysis. It is known as DeMoivre's theorem, after its discoverer, and may be stated thus: The argument of the nth power of any complex number is equal to n times the argument of the number. The theorem may be shown to hold for any value of n, negative, fractional, irrational or even imaginary, but unless n is integral cos n0+ i sin no represents but one of the several values which (cos + i sin 0)" may have. To illustrate the use of DeMoivre's theorem, we will employ it to i Ι i raise +√3 to the 9th power. 2 2 Changing +√3 to the 2 =, sine = (+1√3) = (cos + i sin Similarly π 3 5π ] (1+i)= [√2 (cos + i sin )]-4[cos + sin 4 = −4 4 (1 + i). 152. Roots of Complex Numbers. Suppose it is required to find the nth root of the complex number z = r (cos + i sin 0). Denote the root by z' r' (cose' + i sin '). =r (cos + i sin 0), Then (z')" =z, and we have by DeMoivre's theorem = r' = "r, nơ′ = 0, 0 + 2 π, 0 + 4π, 0 + 6π, . . . 0 + 2kπ, since when an angle is increased by any number of times 2 π both the sine and cosine remain unchanged. It follows that DeMoivre (1667-1754) created a large part of that portion of trigonometry which deals with complex numbers. His death has a curious psychological interest. Shortly before his death he slept a little longer each day, until when the limit of twenty-four hours was reached, he died in his sleep. These values are not all different, for when k = n we have 'cos 0 + 2 nπ n and similarly COS + i sin + 2 nr) = r2 [(cos(+2)+ i sin (2+2)] n COS 0+2(n+1)+isin n n r2 (cos + 2(n + μ)x + i sin® + 2(m + μr) n n 0 + 2 μπ n so that z' has only n distinct values corresponding to the values Every complex number r (cos + i sin 0) has n nth roots given by the formula EXAMPLE I To find the fourth roots of I. Solution. In the trigonometric form = cos o° + i sin o°, hence Denoting the four roots by zo, Z1, Z2, Z3 respectively, we have cos 15 = cos(2-3)=cos 35, sin 1-sin(2x-3)=- sin 37, 5 and COS 5 7 5 cos 2 cos(2-)-cos, sin 2-sin(2x-5)=-sin 5 = so that finally 5 3′′, 5 |