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4-1.Complex numbers
easy
અસમતા $|z - 4|\, < \,|\,z - 2|$ એ . . . ભાગ દર્શાવે છે .
A
${\mathop{\rm Re}\nolimits} (z) > 0$
B
${\mathop{\rm Re}\nolimits} (z) < 0$
C
${\mathop{\rm Re}\nolimits} (z) > 2$
D
એકપણ નહીં.
(AIEEE-2002) (IIT-1982)
Solution
(d) Given inequality $|z – 4|\, < \,|z – 2|$
$ \Rightarrow $ $|z – 4{|^2} < \,|z – 2{|^2} \Rightarrow {(x – 4)^2} + {y^2} < {(x – 2)^2} + {y^2}$
==> $4x > 12 \Rightarrow {\mathop{\rm Re}\nolimits} (z) > 3$.
Standard 11
Mathematics