Q29 / 75 · JEE Main 2020 (09 Jan Shift 2) · Mathematics

If $z$ is a complex number satisfying $|R e (z)| + |I m (z)| = 4,$ then $|z|$ cannot be

A
$\sqrt{\frac{17}{2}}$
B
$\sqrt{10}$
Correct answer
C
$\sqrt{7}$
D
$\sqrt{8}$

Solution

Let $z = x + i y$
i.e. $|x| + |y| = 4$

Using perpendicular distance of line from origin, minimum value of $|z| = 2 \sqrt{2}$
Maximum value of $|z| = 4$
So, $|z|$ can’t be $\sqrt{7}$

Hence, the correct answer is option: 'C'.

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