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1) Solve the following system of equations: −3y 2 + 2xy + x + 9 = 0 x + 4y + 9 = 0 (1) (2) There is always more than one way to do things, but what I would do is notice that there is an x in each equation given. Subtract (ii) from (i) to get −3y 2 + 2xy − 4y = 0. What can you do with this? Factor — every term has a y in it. Get y(−3y + 2x − 4) = 0. Use the fact that this implies either y = 0 or −3x + 2x − 4 = 0. • If y = 0, plug that in to either original equation to get x = −9. • If −3y + 2x − 4 = 0, make a system out of that and equation (ii) above. Use substitution or elimination to solve this: get the solution x = −1, y = −2. Make sure that you present your solutions appropriately: you can’t mix and match the x and y solutions you get... 2) Draw a graph of the following system of inequalities: x2 + y 2 x+y ≥ 9 ≤ 3 The first equation gives a circle centered at the origin of radius 3. The second equation gives a line through the points (0, 3) and (3, 0). The shaded area is the one that lies below and to the left of both curves. 1 3) Given the complex number z = √ 3 − i, find the following: (a) |z| For any complex number x + iy, |z|2 = x2 + y 2 . In this case, then, |z| = 2. (b) z 7 (leave your answer in polar form) Use DeMoivre’s Theorem. First express z in polar form (z = 2(cos(−30 o ) + i sin(−30o ))). Then |z|7 = [2(cos(−30o ) + i sin(−30o ))]7 = 27 (cos(−210o ) + i sin(−210o)) √ 1 − 3 +i ) = 128( 2 √2 = −64 3 + i64 (c) z 1/2 DeMoivre’s Theorem again: |z|1/2 π π = [2(cos(− + 2πk) + i sin(− + 2πk))]1/2 6 6 √ π π = 2(cos(− + πk) + i sin(− + πk)) 12 12 Use the values k = 0, 1 to find the exact answers. On the midterm, be prepared for slightly nicer numbers and the necessity of evaluating completely (to get an answer that looks like x + iy). 4) What is the value of i101 ? You can use DeMoivre’s Theorem, or notice that since i4 = 1, and 101 has remainder 1 when you divide by 4, i101 = i1 = i. A lot of people were off by one on this — not a bad thing to review a little bit. 2