By A.H. Pavel

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How much is the bill? Chapter 6 Complex Numbers We all know that there are simple quadratic equations, such as x2 + 1 = 0, that have no real solutions. In order to provide a notation with which to discuss such equations, we introduce a symbol i, and define i2 = −1 . A complex number is defined to be a symbol a + bi, where a, b are real numbers. If z = a + bi, we call a the real part of z and b the imaginary part, and write a = Re(z), b = Im(z) . We define addition and multiplication of complex numbers by the rules addition: (a + bi) + (c + di) = a + c + (b + d)i multiplication: (a + bi)(c + di) = ac − bd + (ad + bc)i .

Which one is nearest to i in the Argand diagram? √ (c) Find the seven roots of the equation z7 − 3 + i = 0. Which one of these roots is closest to the imaginary axis? 5. Let z be a non-zero complex number. Prove that the three cube roots of z are the corners of an equilateral triangle in the Argand diagram. COMPLEX NUMBERS 47 6. Express √1+i in the form x + iy, where x, y ∈ R. By writing each of 1 + i 3+i √ and 3 + i in polar form, deduce that √ √ 3+1 3−1 π π cos = √ , sin = √ . 12 12 2 2 2 2 7.

Also c2 + s2 = cos2 θ + sin2 θ = 1, so s2 = 1 − c2 , and therefore cos 3θ = c3 − 3c 1 − c2 = 4c3 − 3c . cos 3θ = 4 cos3 θ − 3 cos θ . 5 We now use the previous example to find a cubic equation having cos π9 as a root. 4 gives cos 3θ = 4c3 − 3c . However, cos 3θ = cos π3 = 12 . Hence cos π9 is a root of the cubic equation 1 2 = 4c3 − 3c. In other words, c = 8x3 − 6x − 1 = 0 . Note that if φ = π9 + 23π or π9 + 43π , then cos 3φ = 12 , and hence the above argument shows cos φ is also a root of this cubic equation.

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