MHT CET · Maths · Complex Number
If \(\mathrm{z}=\mathrm{x}+\) iy satisfies the condition \(|\mathrm{z}+1|=1\), then \(\mathrm{z}\) 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
Answer & Solution
Correct Answer
(B) circle with centre \((-1,0)\) and radius 1
Step-by-step Solution
Detailed explanation
Let \(\mathrm{z}=\mathrm{x}+\) iy and \(|\mathrm{z}+1|=1\)
\(
\begin{aligned}
& \therefore|(\mathrm{x}+1)+\mathrm{iy}|=1 \\
& \therefore \sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=1 \Rightarrow(\mathrm{x}+1)^2+\mathrm{y}^2=1
\end{aligned}
\)
This is an equation of a circle with centre \((-1,0)\) and radius 1 .
\(
\begin{aligned}
& \therefore|(\mathrm{x}+1)+\mathrm{iy}|=1 \\
& \therefore \sqrt{(\mathrm{x}+1)^2+\mathrm{y}^2}=1 \Rightarrow(\mathrm{x}+1)^2+\mathrm{y}^2=1
\end{aligned}
\)
This is an equation of a circle with centre \((-1,0)\) and radius 1 .
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