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zimovet [89]
2 years ago
12

Need the answer for this thanks. I got C but that is incorrect​. Please show work thanks! ​

Mathematics
1 answer:
natulia [17]2 years ago
8 0

Answer:

Step-by-step explanation:

a: Wrong. The first thing that you have to notice is that the sum goes to infinity. If you want k=4 to be the last condition, then take out the 3 dots.

b: That's the answer.

c: wrong. You get a real mess when you let set k = 0. Try it on your calculator.

1 ÷ 0 = Watch carefully as your calculator mentally melts down.

d: wrong. It's just not right. The highest power is k^2. There is no way to get k^3

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C

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We have: (I rewrote the function)

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The first iterate will be:

\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}

The second iterate will be:

\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}

And the third iterate will be:

\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}

Hence, our answer is C.

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3 years ago
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