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: Understanding the Question:
The question asks us to identify the geometric locus of the complex number $z = x + iy$ given the modulus condition $|z + 1| = 1$.

Step 2: Key Formula or Approach:
1. Substitute $z = x + iy$ into the given equation.
2. The modulus of any complex number $w = a + ib$ is defined as $|w| = \sqrt{a^2 + b^2}$.
3. Square both sides to eliminate the radical and match it to the standard equation of a circle: $$(x - h)^2 + (y - k)^2 = r^2$$ where $(h,k)$ is the center and $r$ is the radius.

Step 3: Detailed Explanation:
Given the condition: $$|z + 1| = 1$$ Substitute $z = x + iy$: $$|(x + iy) + 1| = 1$$ Group the real and imaginary parts together inside the modulus brackets: $$|(x + 1) + iy| = 1$$ Apply the geometric definition of the modulus: $$\sqrt{(x + 1)^2 + y^2} = 1$$ Square both sides of the equation to eliminate the radical: $$(x + 1)^2 + y^2 = 1^2$$ We can rewrite this expression in the standard circle format: $$(x - (-1))^2 + (y - 0)^2 = 1^2$$ By direct inspection, this matches a circle equation where: The coordinates of the center $(h, k)$ are $(-1, 0)$. The radius $r$ is equal to 1.

Step 4: Final Answer:
The locus of $z$ is a circle with centre $(-1,0)$ and radius 1, which corresponds to option (B).