Complex number $1$

Algebra Level 5

$z$ is a complex number satisfying the equation $(z+1)^5=32z^5$ . It can be shown that all the roots of the above polynomial in $z$ are equidistant from a fixed complex number $x$ .Then find the value of $|x|$ .

The answer is 0.3333.

1 solution

Archit Tripathi
Dec 6, 2016

It is given that $(z + 1)^{5} = 32z^{5}$

Taking modulus of both the sides

$|z + 1|^{5} = 32|z|^{5}$

$\Rightarrow$ $|z + 1| = 2|z|$

$\Rightarrow$ $\frac{|z + 1|}{|z|} = 2$

which represents a circle in the complex plane with centre at $x = \frac{1}{3} +0i$

$\Rightarrow$ $|x| = \frac{1}{3} = \boxed{0.33}$

Exactly the intended solution. Perhaps the locus of $z$ satisfying $|z+1|$ $=$ $2|z|$ being the Apollonius circle is an interesting geometric fact which inspired me to post this problem.

Indraneel Mukhopadhyaya - 4 years, 6 months ago

Complex number 1 1 1

The answer is 0.3333.

1 solution

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Complex number $1$