...(ad + b с)2 [657, Sсh.] о8 с* = norm (а + bi) multiplied by norm (c + di). Cor. — The modulus of the product of two complex numbers equals the product of their moduli. 659. 4. If a + bi = 0, then a = 0 and 5 = 0. For, if a + b г - = 0, bi — — a and — 6s...

...[cos^j + 02) + г sin (^ + 02)]. Figure 71 illustrates the multiplication of 5 — 2 г by 2 + Si. The product is shown to be 16 + 11 i. We have then...of the product equals the sum of their amplitudes. The inverse process of division is readily performed, with the . Т t or D1 - rt (cos в1 + i sin flj)...

...71 illustrates the multiplication of 5 — 2 г by 2 + 3 ¿. The product is shown to be 16 + 11 г. We have then the law that the absolute value of the...of division is readily performed, with the result or г» = 1l [cos (0l - 02) + i sin (0l - 02)] The absolute value of the quotient is equal to the quotient...

...cos a — sin в sin a = cos (в + a), COS в sin a + sin в COS a = sin (0 + a). Hence the modulus of the product of two complex numbers equals the product of their moduli, and the amplitude of the product equals the sum of their amplitudes. The product may be found...

...в cos a — sin в sin a = cos (0 + a), cos в sin a + sin 0 coS a = sin (0 + a). Hence the modulus of the product of two complex numbers equals the product of their moduli, and the amplitude of the product equals the sum of their amplitudes. The product may be found...

...p)l. 1t follows that |aß| = rp = |аiiß|, arg (а0) = 0 + ii - arg а + arg ß, (13) ie, the modulus of the product of two complex numbers equals the product of their moduli, while the argument of the product of two complex numbers equals the sum of their arguments....

...(1-18) (rs)t = r(st) and (9 + </>)+ a = 6 + (</> + a). Again by (1.16) and 1.2.1(a), since that is, the absolute value of the product of two complex numbers equals the product of their absolute values. Also, zw = ~zw (1.20) + 0) = rscis[-(6 + 4>)\ by 1.2.1 (d) = [r cis (-0)][s cis (-<£)] by (1.16) =...