First slide

Geometry of complex numbers

Question

$If three exist a complex number Z satisfying both | Z - m i | = m + 5 and | Z - 4 | < 3, then$

Difficult

A

$m can be - 2$
B

the number of integral value so m is 7
C

$m \in [3, 7]$
D

the number of integral value so m is 5

Solution

$|z - m i| = m + 5 r e p r e s e n t s a c i r c l e h a v i n g c e n t r e a t m i a n d r a d i u s m + 5$
$a n d |z - 4| < 3 i s i n s i d e r e g i o n o f a c i r c l e h a v i n g c e n t r e a t (4, 0) a n d r a d i u s 3$
$T h e t w o c i r c l e s i n t e r s e c t$
$\begin{array}{l} t h e r e f o r e r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2} \\ \Rightarrow m + 5 - 3 < \sqrt{m^{2} + 16} < m + 5 + 3 \\ \Rightarrow m \in (- 3, 3) \end{array}$

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