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  • Help me please, - Complex numbers and quadratic equations - JEE Main-10

Let z= \left ( 2+3i \right )a+\left ( 3-4i \right )b+\left ( i-5 \right ), where  a,b\; \epsilon \; R  is such that \left | z \right |= 0  then a+b equals

  • Option 1)

    2

  • Option 2)

    3

  • Option 3)

    4

  • Option 4)

    5

 

Answers (1)

best_answer

\because \left |z \right |=0\: so\: z=0

\therefore \left ( 2+3i \right )a+\left ( 3-4i \right )b+\left ( i-5 \right )=0

\Rightarrow \: \left ( 2a+3b-5 \right )+i\left ( 3a-4b+1 \right )=0

it is possible when 2a+3b=5 and 3a-4b=-1

Solving this equations we get a=1 and b=1

\therefore Option (A)

 

Property of Modulus of z(Complex Number) -

|z|=0 iff z=0

- wherein

|z|= Modulus of z

 

 


Option 1)

2

This is correct

Option 2)

3

This is incorrect

Option 3)

4

This is incorrect

Option 4)

5

This is incorrect

Posted by

Himanshu

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