Question

If $ |z - 1| $ + $ |z + 3| = 8,$ then the range of value of | $ z - 4| $ is

• A. $ (0,8) $
• B. $ [0, 8 ] $
• C. $ [1,9] $
• D. $ [5, 9] $

Answer 1

Answer: (C)

We have, $|z-1|+|z+3|=8$
$\Rightarrow|z-(1+0 i)|+|z-(-3+0 i)|=8$
Thus, if $P, S$ and $S^{\prime}$ are three points in the argand plane representing complex numbers $z, 1+0 i$ and $-3+0 i$, then from Eq. (i), we have
$P S+P S^{\prime}=8$
$\Rightarrow P$ lies on the ellipse whose two foci are at $S(1,0)$ and $S^{\prime}(-3,0)$ and major axis $=8$

Now, $P Q=|z-4|$
Clearly, $P Q$ is minimum or maximum according as $P$ coincides with $A(3,0)$ and $A^{\prime \prime}(-5,0)$, respectively.
Thus, $P Q=|z-4|$ varies between $A Q=1$ and $A^{\prime} Q=9$
Hence, $|z-4| \in[1,9]$