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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Complex Number

BITSAT 2024 · Mathematics · Complex Number

The modulus of the complex number z such that |z+3-i|=1 and (z)=π...

The modulus of the complex number zz such that ∣z+3−i∣=1|z+3-i|=1 and arg⁡(z)=π\arg (z)=\pi is equal to

Options

  1. A3
  2. B2
  3. C9
  4. D4

Correct answer

A. 3

Step-by-step solution

Let z=x+iyz=x+i y Given, ∣z+3−i∣=1\quad|z+3-i|=1⇒∣(x+3)+(y−1)i∣=1⇒(x+3)2+(y−1)2=1⇒(x+3)2+(y−1)2=1…(i)\begin{aligned} & \Rightarrow \quad|(x+3)+(y-1) i|=1 \\ & \Rightarrow \quad \sqrt{(x+3)^2+(y-1)^2}=1 \\ & \Rightarrow \quad(x+3)^2+(y-1)^2=1\ldots(i) \end{aligned}Above is the equation of circle with centre $\equiv(-3,1) and \text{ radius }=1$ Also, $\arg (z)=\pi \Rightarrow \tan ^{-1}\left|\frac{y}{x}\right|=\pi\Rightarrow \quad\left|\frac{y}{x}\right|=\tan \pi=0 \Rightarrow y=0$ By Eq. (i),\begin{aligned} \Rightarrow & x & =-3 \\ \therefore & z & =-3+0 i \\ \Rightarrow & \text{ Modulus\text{ of } } z & =3 \end{aligned}$$
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