Question

If $z = x + iy$ satisfies the condition $|z + 1| = 1$, then $z$ lies on the

• A. parabola with vertex $(0,0)$
• B. circle with centre $(-1,0)$ and radius 1
• C. circle with centre $(1,0)$ and radius 1
• D. Y-axis

Hint:

In the complex plane, the expression $|z - z_0| = r$ directly represents the geometric locus of a circle whose center is located at the complex coordinate $z_0$ and whose radius is $r$. Rewriting the prompt's expression as $|z - (-1)| = 1$ instantly reveals the center to be $-1$ (or $(-1,0)$ in Cartesian coordinates) and the radius to be $1$.

Answer 1

The Correct Option is B

Step 1: Interpret the modulus geometrically.
The expression $|z + 1| = |z - (-1)|$ is the distance from the point $z$ to the fixed point $-1$ (that is, $(-1, 0)$) in the complex plane.
Step 2: Recognise the locus.
A point that stays a constant distance from a fixed point traces a circle. So $z$ lies on a circle centred at $(-1, 0)$.
Step 3: Read the radius.
Here the constant distance is 1, so the radius is 1.
Step 4: Confirm algebraically.
Put $z = x + iy$: $|(x+1) + iy| = 1$ means $\sqrt{(x+1)^2 + y^2} = 1$.
Step 5: Square both sides.
$(x+1)^2 + y^2 = 1$, i.e. $(x-(-1))^2 + (y-0)^2 = 1^2$.
Step 6: Match to the standard circle.
Comparing with $(x-h)^2 + (y-k)^2 = r^2$ gives centre $(-1, 0)$ and radius $1$.
\[ \boxed{\text{circle with centre } (-1, 0) \text{ and radius } 1} \]