MHT CET · Maths · Complex Number
If amplitude of \((z-2-3 i)\) is \(\frac{3 \pi}{4}\), then locus of \(z\) is (where \(\mathrm{z}=\mathrm{x}+\) iy)
- A \(x+y=1\)
- B \(x+y=5\)
- C \(x-y=-5\)
- D \(x-y=1\)
Answer & Solution
Correct Answer
(B) \(x+y=5\)
Step-by-step Solution
Detailed explanation
Amplitude of \((z-2-3 i)\) is \(\frac{3 \pi}{4}\) and we have \(z=x+\) iy
\(\therefore \operatorname{Amp}[(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-3)]\) is \(\frac{3 \pi}{4}\)
\(\therefore \tan ^{-1}\left(\frac{\mathrm{y}-3}{\mathrm{x}-2}\right)=\frac{3 \pi}{4} \Rightarrow \tan \left(\frac{3 \pi}{4}\right)=\frac{\mathrm{y}-3}{\mathrm{x}-2}\)
\(\therefore-1=\frac{y-3}{x-2} \Rightarrow-x+2=y-3 \Rightarrow x+y=5\)
\(\therefore \operatorname{Amp}[(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-3)]\) is \(\frac{3 \pi}{4}\)
\(\therefore \tan ^{-1}\left(\frac{\mathrm{y}-3}{\mathrm{x}-2}\right)=\frac{3 \pi}{4} \Rightarrow \tan \left(\frac{3 \pi}{4}\right)=\frac{\mathrm{y}-3}{\mathrm{x}-2}\)
\(\therefore-1=\frac{y-3}{x-2} \Rightarrow-x+2=y-3 \Rightarrow x+y=5\)
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