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Chapter 2: Q15P (page 59)

Find the disk on convergence for each of the following complex power series.
${鈭�}_{n - 0}^{m} \frac{{(z - 2 + i)}^{n}}{2^{n}}$

Short Answer

Expert verified

The region of convergent is $|z - 2 + i| < 2$ .

Step by step solution

01

Given data

The given series is, ${鈭�}_{n = 0}^{鈭�} \frac{{(z - 2 + i)}^{n}}{2^{n}}$ .

02

Concept of Ration test

The ratio test is a check (or "criterion") to see if a series will eventually converge to ${鈭�}_{n - 1}^{鈭�} a_{n}$ . Every term is a real or complex number, and when n is big, is not zero.

03

Calculation to check the series is convergent

Find $A_{n}$ and $A_{n + 1}$ as follows:

$A_{n} = \frac{{(z - 2 + i)}^{n}}{2^{n}} A_{n + 1} = \frac{{(z - 2 + i)}^{n + 1}}{2^{n + 1}}$

Apply ratio test as follows:

role="math" localid="1658727517111" $蚁 = \underset{n 鈫� 鈭�}{l i m} |\frac{A_{n + 1}}{A_{n}}| 蚁 = \underset{n 鈫� 鈭�}{l i m} |\frac{\frac{(2 - 2 + i^{n + 1}}{2^{n + 1}}}{\frac{{(2 - 2 + i)}^{n}}{2^{n}}}| 蚁 = |\frac{(z - 2 + i)}{2}|$

If, $蚁 < 1$ , then the series is convergence.

04

Calculation to find the region of convergent

The region of convergence is given as follows:

$|\frac{(z - 2 + i)}{2}| < 1 |z - 2 + i| < 2$

Let, $z = x + i y$ .

Therefore, calculate as follows:

$|x + i y - 2| < 2 |(x - 2) + (y + 1) i| < 2 \sqrt{{(x - 2)}^{2} + {(y + 1)}^{2}} < 2 {(x - 2)}^{2} + {(y + 1)}^{2} < 2$

It is an equation of disk with center $(2, - 1)$ and $r = 2$ .

Hence the region of convergent is $|z - 2 + i| < 2$ .

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Most popular questions from this chapter

Solve for all possible values of the real numbers x and y in the following equations. $| 1 - (x + i y) | = x + i y$

Question: Find vand a if $z = c o s 2 t + i s i n 2 t$ , can you describe the motion.

Find the absolute value of each of the following using the discussion above. Try to do simple problems like these in your head-it saves time.

$\frac{5 - 2 i}{5 + 2 i}$ .

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