Question

Among the complex number z satisfying condition $|z+1-i|\le 1,$the number having the least positive argument is

• A. $1-i$
• B. $-1+i$
• C. $-i$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Finding the number having the least positive argument:

$|z-(-1+i\left)\right|\le 1$ represents a region of a circle with centre $(-1+i)$ and radius $1$. The point on this circle with least amplitude is $P(0,1)$ and corresponds to $i$.

Given $\begin{array}{rcl}|z+1-i|& \le & 1...\left(1\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & z=1-i\end{array}$

Substitute in (1)

$\Rightarrow \mid 1-i+1-i\mid \nleqq 1$

Does not satisfy

Put $z=-1+i$ in (1)

$\left|-1+i+1-i\right|\le 1$

Argument of $z=\frac{-\mathrm{\pi }}{4}$ $\left[\mathbf{\because }\mathit{a}\mathit{r}\mathit{g}\mathbf{}\mathit{z}\mathbf{}\left(\theta \right)\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{realpart}{imaginarypart}\right)\right]$

Put$z=-i$

$\mid -i+1-i\mid \nleqq 1$

Does not satisfy

Hence, option (D) is the answer.