16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T499,{"@context":"https://schema.org","@type":"Quiz","name":"Bitsat 2024 Complex Numbers PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/bitsat/questions/2024/mathematics/complex-numbers/m7pupupd","educationalLevel":"Entrance exam preparation","about":"Bitsat / Mathematics / Complex Numbers","hasPart":{"@type":"Question","name":"Bitsat 2024 Complex Numbers PYQ with Solution","text":"The modulus of the complex number z such that |z+3-i|=1 and \\arg (z)=\\pi is equal to","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"Let z=x+i y Given, |z+3-i|=1 \\begin{array}{lr} \\Rightarrow & |(x+3)+(y-1) i|=1 \\\\ \\Rightarrow & \\sqrt{(x+3)^{2}+(y-1)^{2}}=1 \\\\ \\Rightarrow & (x+3)^{2}+(y-1)^{2}=1 \\end{array} \\quad \\ldots \\text { (i) } Above is the equation of circle with centre \\equiv(-3,1) and 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{array}{lrl} \\Rightarrow & x & =-3 \\\\ \\therefore & z & =-3+0 i \\\\..."}}}18:T464,

Let $ z=x+i y $

Given,

$ |z+3-i|=1 $

$ \begin{array}{lr} \Rightarrow & |(x+3)+(y-1) i|=1 \\ \Rightarrow & \sqrt{(x+3)^{2}+(y-1)^{2}}=1 \\ \Rightarrow & (x+3)^{2}+(y-1)^{2}=1 \end{array} \quad \ldots \text { (i) } $

Above is the equation of circle with centre $ \equiv(-3,1) $ and

radius $ =1 $.

Also, $ \arg (z)=

Complex Numbers

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