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Q. If `z_1` and `z_2` are two complex numbers such that `|z_1 - z_2|=||z_1|-|z_2||`...

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted August 21, 2013 at 6:26 AM via web

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Q. If `z_1` and `z_2` are two complex numbers such that `|z_1 - z_2|=||z_1|-|z_2||`

then `argz_1 - argz_2` is equal to

A) `-pi/4`

B) `-pi/2`

C) `pi/2`

D) 0

1 Answer | Add Yours

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 21, 2013 at 7:18 AM (Answer #1)

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`z_1` and `z_2` are two complex numbers. Let` z_1 = x_1 + i*y_1` and `z_2 = x_2 + i*y_2` .

`|z_1 - z_2| = sqrt((x_1- x_2)^2 + (y_1 - y_2)^2)`

`|(|z_1| - |z_2|)| = |sqrt((x_1)^2+(y_1)^2) - sqrt((x_2)^2+(y_2)^2)|`

`sqrt((x_1- x_2)^2 + (y_1 - y_2)^2)` = `|sqrt((x_1)^2+(y_1)^2) - sqrt((x_2)^2+(y_2)^2)|`

Take the square of both the sides.

=> `(x_1- x_2)^2` + `(y_1 - y_2)^2` = `(x_1)^2` +`(y_1)^2` + `(x_2)^2` +`(y_2)^2` - `2*sqrt(((x_1)^2+(y_1)^2)((x_2)^2+(y_2)^2))`

=> `x_1*x_2` + `y_1*y_2` = `2*sqrt(((x_1)^2+(y_1)^2)((x_2)^2+(y_2)^2))`

=> `2*x_1*x_2*y_1*y_2` = `(x_1)^2*(y_2)^2 + (x_2)^2*(y_1)^2`

=> `(y_1*x_2 - x_1*y_2)^2 = 0`

=> `y_1*x_2 - x_1*y_2 = 0`

=> `y_1/x_1 = y_2/x_2`

The value of `arg(z_1) - arg(z_2)` is 0.

The correct answer is option D.

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