Complex Area

Geometry Level 5

For a given sequence of real constants, $\displaystyle \left\{\theta _i \right\}_{i=0}^n \text{, } n \in \mathbb{N} \text{ and } n \geq 2$ , the following equation holds true,

$\displaystyle \sum_{r=0}^n z^r \cos \theta _{n-r} = 2$

where $z$ is a complex number.

Given that $\displaystyle |z| < 1$ and $A$ denotes the minimum area in the argand plane in which the roots of the above equation lie, concluded solely by the above information, then

Evaluate : $\displaystyle \lfloor 100A \rfloor$

The answer is 235.

1 solution

Sudeep Salgia
Jun 12, 2014

This question incorporates a different use of Triangle Inequality.

By Triangle Inequality,
$\displaystyle | \sum_{r=0}^n z^r \cos \theta _{n-r} | \leq \sum_{r=0}^n |z^r \cos \theta _{n-r}|$
$\displaystyle \Rightarrow | \sum_{r=0}^n z^r \cos \theta _{n-r} | \leq \sum_{r=0}^n |z^r| | \cos \theta _{n-r}| \leq \sum_{r=0}^n |z|^r (1)$

$\displaystyle \Rightarrow | \sum_{r=0}^n z^r \cos \theta _{n-r} | \leq \frac{ 1 - |z|^{n+1} }{1 - |z| } < \frac{1}{1 - |z| }$

Hence,
$\displaystyle \frac{1}{1 - |z| } > 2 \text{ } \Rightarrow |z| > \frac{1}{2} \text{ since, } |z| < 1$ .

Therefore, $\displaystyle \frac{1}{2} < |z| < 1$ . It represents a disc on the Argand plane with the inner and outer radii as $\frac{1}{2}$ and $1$ respectively. $\displaystyle \Rightarrow A = \frac{3\pi }{4}$ .

$\displaystyle \Rightarrow \boxed{ \lfloor 100A \rfloor = 235}$

Awesome problem!

Ishan Singh - 6 years, 12 months ago

Complex Area

The answer is 235.

1 solution

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