Among the complex number z satisfying the condition $\left| {z + 1 - i} \right| \leqslant 1$
A. 1 – i
B. –1 + i
C. –i
D. None of these

Hint: Here in this question, we have to check for the modulus of the complex number. So, we will check each of the option by putting in the expression $\left| {z + 1 - i} \right|$ and then calculate its modulus and finally check whether it follows the given inequality or not.

Complete step-by-step answer:
We need to check among the given options that which of the given complex numbers has its magnitude less than or equal to 1.
So we’ll check by putting the value of each option in our given equation.
 it is given to us that
$\left| {z + 1 - i} \right| \leqslant 1$ …… (1)
For each of the option we will calculate$\left| {z + 1 - i} \right|$.
We know that for a complex number x + iy, the magnitude is given as:
$\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $.
(1) Let us first put Z = 1- i
On substituting z in $\left| {z + 1 - i} \right|$, we have
$\left| {1 - i + 1 - i} \right| = \left| {2 - 2i} \right| = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 $, which is greater than 1.
So it does not satisfy equation 1.
(2) Let us first put Z = -1+ i
On substituting z in$\left| {z + 1 - i} \right|$, we have
$\left| { - 1 + i + 1 - i} \right| = \left| 0 \right| = 0$, which is less than 1.
So it satisfies the equation 1.
(3) Let us first put Z = - i
On substituting z in $\left| {z + 1 - i} \right|$, we have
$\left| { - i + 1 - i} \right| = \left| {1 - 2i} \right| = \sqrt {{1^2} + {2^2}} = \sqrt 5 $, which is greater than 1.
So it does not satisfy equation 1.

So, the correct answer is “Option B”.

Note: In this type of question, you should remember the formula to calculate the magnitude of the given complex number. The modulus of a complex number z = x + iy is given as $\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} $and its argument $\theta $ is given as ${\tan ^{ - 1}}\dfrac{y}{x}$. Geometrically, $\left| z \right|$ represents the distance of the point ‘z’ in a complex plane from the origin.