Given that $| z - 1 | = 1$ , where z is a point on the Argand plane. Find $\frac{z - 2}{z}$

- i t a n (a r g z)

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t a n (a r g z)

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- t a n (a r g z)

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i t a n (a r g z)

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Solution

The correct option is D $i t a n (a r g z)$
We have
$| z - 1 | = 1$
Let $z - 1 = c o s θ + i s i n θ$
$∴ z - 2 = c o s θ + i s i n θ - 1$
$= - 2 s i n^{2} θ / 2 + 2 i s i n θ / 2 c o s θ / 2$
$z - 2 = 2 i s i n θ / 2 (c o s θ / 2 + i s i n θ / 2)$ ........ (1)
and $z = 1 + c o s θ + i s i n θ$
$= 2 c o s^{2} θ / 2 + 2 i s i n θ / 2 c o s θ / 2$
$z = 2 c o s θ / 2 (c o s θ / 2 + i s i n θ / 2)$ ....... (2)
from (1) and (2)
$\frac{z - 2}{z} = i t a n θ / 2$
$= i t a n (a r g z)$ ( $∵ a r g z = θ / 2$ from (2))
Ans: D

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