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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Complex Number

If z is a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z|...

If zz is a complex number satisfying \left|Re\left(z\right)\right| + \left|Im\left(z\right)\right| = 4,then∣z∣then \left|z\right| cannot beIf z is a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be

Options

  1. A172\sqrt{\dfrac{17}{2}}
  2. B10\sqrt{10}
  3. C7\sqrt{7}
  4. D8\sqrt{8}

Correct answer

C. 7\sqrt{7}

Step-by-step solution

Let z=x+iyz = x + iy
i.e. \left|x\right| + \left|y\right| = 4</p><p><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/846c6b61−9c3a−4699−a0b8−58c380a00218.jpeg"/><br/>Usingperpendiculardistanceoflinefromorigin,minimumvalueof</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/846c6b61-9c3a-4699-a0b8-58c380a00218.jpeg" /><br />Using perpendicular distance of line from origin, minimum value of\left|z\right| = 2\sqrt{2}<br/>Maximumvalueof∣z∣=4<br />Maximum value of \left|z\right| = 4
So, \left|z\right|can&rsquo;t be\sqrt{7}$

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