lf the roots of $(z - 1)^{n} = 2 w (z + 1)^{n}$ where $n \geq 3$ and $w$ is a complex cube root of unity, are plotted in the argand plane, then these roots lie on

straight line

No worries! We‘ve got your back. Try BYJU‘S free classes today!

circle

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

ellipse

No worries! We‘ve got your back. Try BYJU‘S free classes today!

rectangular hyperbola

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A circle
$(z - 1)^{n} = 2 w (z + 1)^{n}$
$(\frac{z - 1}{z + 1})^{n} = 2 w$
$| \frac{z - 1}{z + 1} |^{n} = | 2 w | = 2 | w |$
$| \frac{z - 1}{z + 1} |^{n} = 2$
$| \frac{z - 1}{z + 1} | = 2^{\frac{1}{n}}$
Now $2^{\frac{1}{n}} > 1$ for $n ϵ N$
Therefore, $z$ lies on a circle.
Hence, option 'B' is correct.

Suggest Corrections

Similar questions

{(z - 1)}^{25} = 2 ω^{2} {(z + 1)}^{25} (

ω

)

(z - 1)^{n} = i (z + 1)^{n}

(z - 1)^{n} = i (z + 1)^{n}

z^{2} + z + 1 = 0

n

n \sum k = 1 (z^{k} + z^{- k})^{2} = n + 3 [\frac{n}{3}]

[x]

\leq x

w \neq 1

w^{k} + w^{- k} = {\begin{matrix} 0 - 1 & if k is not a multiple of 3 2 & if k is a multiple of 3 \end{matrix}

z^{n} = (z + 1)^{n}

Join BYJU'S Learning Program