Question

(a) Find all numbers z є C such that (z-i)"--64. (b) Find all z E C such that 22 -224i. (c) Find all z E C such that z + z-1-2 . (d) Simplify the expression 1 e i 2 . That is, find the square of the modulus of the complex number 1-e-28 i

Answer 1

a). (z-i)⁶ = -64 = (±2i)⁶. Therefore, z-i = ±2i. If z-i = 2i, then z = 3i and if z-i = -2i, then z = - i. Thus, either z = 3i or, z = -i.

(b). z²- z^-2 = 4i or, z²-1/z² = 4i or, (z⁴-1) /z² = 4i so that z⁴-1 = 4iz² or, z⁴- 4iz² -1 = 0. Now, let y = z². Then, we have y² -4i y -1 = 0. On using the quadratic formula, we get y = [4i±√{ -16-4*1*(-1)]]/2*1 = (4i±√-12)/2 = (2i±i√3). Then z = ±√y = ±(2i±i√3)^1/2.

(c ). Let z = x+iy. Then, x+iy +1/( x+iy) = 2(x-iy) or, (x+iy)² +1 = 2(x+iy)(x-iy) or, x²+2ixy-y² = 2(x² +y²) or,             3y²-2ixy+x² = 0.On using the quadratic formula, we get y=[2ix±√{-4x²-4*3*x²}]/2*3 =[2ix±√(-16x²)]/6 = (2ix± 4ix)/6 . Thus, either y =(2ix+ 4ix)/6 = ix or, y = (2ix- 4ix)/6 = -ix/3. Then z = x+iy = x+ i(ix) = x-x = 0 or, z = x+i(-ix/3) = x+x/3 = 4x/3 . Thus, either z = 0 or, z = 4x/3, where x is an arbitrary real number.

Please post the remaining part again, separately.