Answer:
a. 5 + 2i, 13 + 2i, 29 + 2i
Step-by-step explanation:
We'll use the formula f(z) = 2z + (3 - 2i) for each iteration. The output of the first iteration will be come the input of the second iteration, and so on.
So, we start with z0 = 1 + 2i and we plug that into the base equation:
z0 = 1 + 2i ==> f(z) = 2(1 + 2i) + 3 - 2i = 2 + 4i + 3 - 2i = 5 + 2i
z1 = 5 + 2i ==> f(z) = 2(5 + 2i) + 3 - 2i = 10 + 4i + 3 - 2i = 13 + 2i
z2 = 13 + 2i ==> f(z) = 2(13 + 2i) + 3 - 2i = 26 + 4i + 3 - 2i = 29 + 2i
z3 = 29 + 2i
<span>A bond that closed today at 94 down 2 closed yesterday in dollars at </span>940 dollars
So the original price is "x".
the discounted price by 10% is P(x) = 0.9x.
the price minus a $150 coupon is C(x) = x - 150.
so, if you go to the store, the item is discounted by 10%, so you're really only getting out of your pocket 90% of that, or 0.9x, but!!! wait a minute!! you have a $150 coupon, and you can use that for the purchase, so you're really only getting out of your pocket 0.9x - 150, namely the discounted by 10% and then the saving from the coupon.
C( P(x) ) = P(x) - 150
C( P(x) ) = 0.9x - 150
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