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Leviafan [203]
1 year ago
5

Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of

Mathematics
2 answers:
nikklg [1K]1 year ago
8 0

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

Lerok [7]1 year ago
3 0

Answer:a

Step-by-step explanation:

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If <u>quadrilaterals</u> WXYZ and BADC are <u>congruent</u>, then their corresponding <u>sides</u> are congruent.

Given that

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patriot [66]

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