Question

Let z = x + iy be a complex number such that $|z+i|=2$. Then the locus of z is a circle whose centre and radius are

• A. (0, -1); 2
• B. (0, 2); 2
• C. (1,-1); 2
• D. $(0,-1); \sqrt{3}$
• E. (0, 2); $\sqrt{3}$

Hint:

Geometry Tip: You can solve this instantly without algebra! The equation $|z - (-i)| = 2$ directly translates to "the distance from $z$ to the point $-i$ is exactly 2." The point $-i$ corresponds to $(0, -1)$ on the coordinate plane.

Answer 1

The Correct Option is A

Concept:
The modulus of a complex number $w = a + ib$ is defined as $|w| = \sqrt{a^2 + b^2}$. The geometric interpretation of $|z - z_0| = r$ is a circle in the complex plane with center at $z_0$ and radius $r$.

Step 1: Substitute algebraic form into the equation.
Given the equation $|z + i| = 2$, replace the complex variable $z$ with its Cartesian form $x + iy$: $$|x + iy + i| = 2$$

Step 2: Group real and imaginary components.
Factor out the imaginary unit '$i$' to clearly separate the real part and imaginary part: $$|x + i(y + 1)| = 2$$ Here, the real part is $x$ and the imaginary part is $(y + 1)$.

Step 3: Apply the modulus definition.
Calculate the magnitude by taking the square root of the sum of the squares of the real and imaginary parts: $$\sqrt{x^2 + (y + 1)^2} = 2$$

Step 4: Square both sides to form standard circle equation.
Remove the square root to match standard Cartesian formats: $$x^2 + (y + 1)^2 = 4$$ This can be rewritten to perfectly match the standard circle equation $(x-h)^2 + (y-k)^2 = r^2$: $$(x - 0)^2 + (y - (-1))^2 = 2^2$$

Step 5: Extract the center and radius.
From the standard form $(x-h)^2 + (y-k)^2 = r^2$, the center is $(h, k)$ and the radius is $r$: $$\text{Center } (h, k) = (0, -1)$$ $$\text{Radius } r = 2$$ Hence the correct answer is (A) (0, -1); 2.