JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 20)
Let z and $$\omega$$ be two complex numbers such that $$\omega = z\overline z - 2z + 2,\left| {{{z + i} \over {z - 3i}}} \right| = 1$$ and Re($$\omega$$) has minimum value. Then, the minimum value of n $$\in$$ N for which $$\omega$$n is real, is equal to ______________.
Answer
4
Explanation
Let z = x + iy
| z + i | = | z $$-$$ 3i |
$$ \Rightarrow $$ y = 1
Now
$$\omega$$ = x2 + y2 $$-$$ 2x $$-$$ 2iy + 2
$$\omega$$ = x2 + 1 $$-$$ 2x $$-$$ 2i + 2
Re($$\omega$$) = x2 $$-$$ 2x + 3
Re($$\omega$$) = (x $$-$$ 1)2 + 2
Re($$\omega$$)min at x = 1 $$ \Rightarrow $$ z = 1 + i
Now,
$$\omega$$ = 1 + 1 $$-$$ 2 $$-$$ 2i + 2
$$\omega$$ = 2(1 $$-$$ i) = 2$$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$$
$$\omega$$n = 2$$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$$
If $$\omega$$n is real $$ \Rightarrow $$ n = 4
| z + i | = | z $$-$$ 3i |
$$ \Rightarrow $$ y = 1
Now
$$\omega$$ = x2 + y2 $$-$$ 2x $$-$$ 2iy + 2
$$\omega$$ = x2 + 1 $$-$$ 2x $$-$$ 2i + 2
Re($$\omega$$) = x2 $$-$$ 2x + 3
Re($$\omega$$) = (x $$-$$ 1)2 + 2
Re($$\omega$$)min at x = 1 $$ \Rightarrow $$ z = 1 + i
Now,
$$\omega$$ = 1 + 1 $$-$$ 2 $$-$$ 2i + 2
$$\omega$$ = 2(1 $$-$$ i) = 2$$\sqrt 2 {e^{i\left( {{{ - \pi } \over 4}} \right)}}$$
$$\omega$$n = 2$$\sqrt 2 {e^{i\left( {{{ - n\pi } \over 4}} \right)}}$$
If $$\omega$$n is real $$ \Rightarrow $$ n = 4
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