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JEE Mains · Maths · STD 11 - 4.1 complex nubers

જો સંકર સંખ્યા \(z\) એ \(|\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4\) નું સમાધાન કરે છે તો \(|z|\) ની કિમંત  . . . શક્ય નથી.

  1. A \(\sqrt{\frac{17}{2}}\)
  2. B \(\sqrt{10}\)
  3. C \(\sqrt{8}\)
  4. D \(\sqrt{7}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt{7}\)

Step-by-step Solution

Detailed explanation

\(z=x+i y\) \(|x|+|y|=4\) \(|z|=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}} \Rightarrow|\mathrm{z}|_{\min }=\sqrt{8} \&|\mathrm{z}|_{\max }=4=\sqrt{16}\) So \(|z|\) cannot be \(\sqrt{7}\)
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