JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 4)
Let z be complex number such that
$$\left| {{{z - i} \over {z + 2i}}} \right| = 1$$ and |z| = $${5 \over 2}$$.
Then the value of |z + 3i| is :
$$\left| {{{z - i} \over {z + 2i}}} \right| = 1$$ and |z| = $${5 \over 2}$$.
Then the value of |z + 3i| is :
$$2\sqrt 3 $$
$$\sqrt {10} $$
$${{15} \over 4}$$
$${7 \over 2}$$
Explanation
Given $$\left| {{{z - i} \over {z + 2i}}} \right| = 1$$
|z – i| = |z + 2i|
(let z = x + iy)
$$ \Rightarrow $$ x2 + (y – 1)2 = x2 + (y + 2)2
$$ \Rightarrow $$ y = $$ - {1 \over 2}$$
Also given |z| = $${5 \over 2}$$
$$ \Rightarrow $$ x2 + y2 = $${{25} \over 4}$$
$$ \Rightarrow $$ x2 = 6
$$ \therefore $$ z = $$ \pm \sqrt 6 $$ - $$ - {1 \over 2}i$$
|z + 3i| = $$\sqrt {6 + {{25} \over 4}} $$ = $${7 \over 2}$$
|z – i| = |z + 2i|
(let z = x + iy)
$$ \Rightarrow $$ x2 + (y – 1)2 = x2 + (y + 2)2
$$ \Rightarrow $$ y = $$ - {1 \over 2}$$
Also given |z| = $${5 \over 2}$$
$$ \Rightarrow $$ x2 + y2 = $${{25} \over 4}$$
$$ \Rightarrow $$ x2 = 6
$$ \therefore $$ z = $$ \pm \sqrt 6 $$ - $$ - {1 \over 2}i$$
|z + 3i| = $$\sqrt {6 + {{25} \over 4}} $$ = $${7 \over 2}$$
Comments (0)
