SATHEE: Complex Numbers 2 Question 38

Complex Numbers 2 Question 38

39.

Prove that there exists no complex number $z$ such that $| z | < 1 / 3$ and $\sum_{r = 1}^{n} a_{r} z^{r} = 1$ , where $| a_{r} | < 2$ .

(2003, 2M)

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Solution:

Given, $a_{1} z + a_{2} z^{2} + \dots + a_{n} z^{n} = 1$

$\begin{array}{lc} and & | z | < \frac{1}{3} \\ ∴ & | a_{1} z + a_{2} z^{2} + a_{3} z^{3} + \dots + a_{n} z^{n} | = 1 \\ \Rightarrow & | a_{1} z | + | a_{2} z^{2} | + | a_{3} z^{3} | + \dots + | a_{n} z^{n} | \geq 1 \end{array}$

[using $| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} |$ ]

$\Rightarrow 2 (| z | + | z |^{2} + | z |^{3} + \dots + | z |^{n}) > 1$

[using $| a_{r} | < 2]$

$\Rightarrow \frac{2 | z | (1 - | z |^{n})}{1 - | z |} > 1$ [using sum of $n$ terms of GP]

$\Rightarrow 2 | z | - 2 | z |^{n + 1} > 1 - | z |$

$\begin{array}{ll} \Rightarrow & 3 | z | > 1 + 2 | z |^{n + 1} \\ \Rightarrow & | z | > \frac{1}{3} + \frac{2}{3} | z |^{n + 1} \\ \Rightarrow & | z | > \frac{1}{3}, which contradicts \end{array}$

$∴$ There exists no complex number $z$ such that

$| z | < 1 / 3 and \sum_{r = 1}^{n} a_{r} z^{r} = 1$

Complex Numbers 2 Question 38

39.

Solution:

Engineering Preparation

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Complex Numbers 2 Question 38

39.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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