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  • Need explanation for: - Complex numbers and quadratic equations - JEE Main-4

Find distance of z= -12-5i from origin.

  • Option 1)

    11 Units

  • Option 2)

    12 Units

  • Option 3)

    13 Units

  • Option 4)

    14 Units

 

Answers (1)

best_answer

\left | z \right |= distance of z from origin

When z=x+iy then \left |z \right |=\sqrt{x^{2}+y^{2}}

here z=-12-5i, So x=-12\: and\: y=-5

So, \left | z \right |=\sqrt{\left ( -12 \right )^{2}+\left ( -5 \right )^{2}}=\sqrt{144+25}=\sqrt{169}=13

\therefore Option (C)

 

Definition of Modulus of z(Complex Number) -

\left | z \right |=\sqrt{a^{2}+b^{2}} is the distance of z from origin in Argand plane

- wherein

Real part of z = Re (z) = a & Imaginary part of z = Im (z) = b

 

 


Option 1)

11 Units

This is incorrect

Option 2)

12 Units

This is incorrect

Option 3)

13 Units

This is correct

Option 4)

14 Units

This is incorrect

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Plabita

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