Question

If |Z₁| = |Z₂| and arg (Z₁) + arg (Z₂) = 0, then

Options

• Z₁ = Z₂

• `Z_1 = barZ_2`

• Z₁Z₂ = 1

• None of these

Answer 1

`Z_1 = barZ_2`

Explanation:

Let `z_1 = x_1 + iy_1` and `z_2 = x_2 + iy_2`

We have given that, `|z_1| = |z_2|`

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

`x_1^2 + y_1^2 = x_2^2 + y_2^2`

`x_1^2[1 + (y_1/x_1)^2] = x_2^2 [1 + (y_2/x_2)^2]`  ......(i)

⇒ Also given that, arg (z₁) + arg (z₂) = 0

`tan^-1 (y_1/x_1) + tan^-1 (y_2/x_2)` = 0

`tan^-1 (y_1/x_1) = - tan^-1 (y_2/x_2)`

`tan^-1 (y_1/x_1) = tan^-1 (y_2/x_2)`

∴ `y_1/x_1 = (-y_2)/(x_2)`  ......(ii)

`y_1^2/x_1^2 = y_2^2/x_2^2 ⇒ (y_1/x_1)^2 = (y_2/x_2)^2`  ......(iii)

From equation (iii) and equation (i), we get

`x_1^2 [1 + (y_1/x_1)^2] = x_2^2 [1 + (y_1/x_1)^2]`

∴ `x_1^2 = x_2^2 ⇒ x_1 = x_2`  ......(iv)

From equation (ii) and equation (iv), we get

∴ `y_1 = - y_2`  ......(v)

By using equation (iv) and equation (v), It can be concluded that `Z_1 = barZ_2` (or) `Z_2 = barZ_1`