Question:

If $z$ be a complex number satisfying $\left|Re\left(z\right)\right|+\left|Im\left(z\right)\right|=4,$ then $|z|$ cannot be :

Updated On: Aug 4, 2024
  • $\sqrt{7}$
  • $\sqrt{\frac{17}{2}}$
  • $\sqrt{10}$
  • $\sqrt{8}$
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The Correct Option is A

Solution and Explanation

$z = x + iy \quad\quad|x| + |y| = 4$
$\left|z\right| = \sqrt{x^{2}+y^{2}} \Rightarrow \left|z\right|_{min} = \sqrt{8} \,\& \,\left|z\right|_{max} = 4 = \sqrt{16}$
So |z| cannot be $\sqrt{7}$
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.