JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 14)
Let z1 and z2 be two complex numbers satisfying | z1 | = 9 and | z2 – 3 – 4i | = 4. Then the minimum value of
| z1 – z2 | is :
0
1
2
$$\sqrt 2 $$
Explanation
$$\left| {{z_1}} \right| = 9,\,\,\left| {{z_2} - \left( {3 + 4i} \right)} \right| = 4$$
$${C_1},(0,0)$$ radius r1 = 9
C2 (3, 4), radius r2 = 4
C1C2 = $$\left| {{r_1} - {r_2}} \right| = 5$$
$$ \therefore $$ Circle touches internally
$$ \therefore $$ $${\left| {{z_1} - {z_2}} \right|_{\min }} = 0$$
$${C_1},(0,0)$$ radius r1 = 9
C2 (3, 4), radius r2 = 4
C1C2 = $$\left| {{r_1} - {r_2}} \right| = 5$$
$$ \therefore $$ Circle touches internally
$$ \therefore $$ $${\left| {{z_1} - {z_2}} \right|_{\min }} = 0$$
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