Here are the solutions. 1. Express z = (40^) - i(40^) in a true polar form. 2. Find the imaginary part of (1)/(1+i). 3. Find (x,y); x,y R, if 3y+2xi = x+i+10. 4. Evaluate the following powers of i: a)

Here are the solutions. 1. Express z = (40^) - i(40^) in a true polar form. The standard polar form is z = r( + i). Given z = (40^) - i(40^). Using (-) = () and (-) = -(): z = (-40^) + i(-40^) The modulus r=1. The argument = -40^. The final answer is z = (-40^) + i(-40^). 2. Find the imaginary part of (1)/(1+i). Multiply by the conjugate of the denominator: (1)/(1+i) = (1)/(1+i) × (1-i)/(1-i) = (1-i)/(1^2 - i^2) = (1-i)/(1 - (-1)) = (1-i)/(2) = (1)/(2) - (1)/(2)i The imaginary part is the coefficient of i. The final answer is -(1)/(2). 3. Find (x,y); x,y R, if 3y+2xi = x+i+10. Equate the real parts and the imaginary parts: Real parts: 3y = x+10 (1) Imaginary parts: 2x = 1 (2) From (2): x = (1)/(2) Substitute x into (1): 3y = (1)/(2) + 10 3y = (1)/(2) + (20)/(2) 3y = (21)/(2) y = (21)/(2 × 3) y = (7)/(2) The final answer is ((1)/(2), (7)/(2)). 4. Evaluate the following powers of i: Recall that i^1=i, i^2=-1, i^3=-i, i^4=1. The powers repeat every 4. *a) i^8: i^8 = (i^4)^2 = 1^2 = 1 *b) i^11: i^11 = i^4 × 2 + 3 = (i^4)^2 · i^3 = 1^2 · (-i) = -i *c) i^42: i^42 = i^4 × 10 + 2 = (i^4)^10 · i^2 = 1^10 · (-1) = -1 *d) i^105: i^105 = i^4 × 26 + 1 = (i^4)^26 · i^1 = 1^26 · i = i The final answers are: a) 1, b) -i, c) -1, d) i. 5. Find the reciprocal of the number z=i. The reciprocal is (1)/(z): (1)/(i) = (1)/(i) × (-i)/(-i) = (-i)/(-i^2) = (-i)/(-(-1)) = (-i)/(1) = -i The final answer is -i.