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प्रश्न
If arg(z – 1) = arg(z + 3i), then find x – 1 : y. where z = x + iy.
उत्तर
Given that: arg(z – 1) = arg(z + 3i)
⇒ arg[x + yi – 1] = arg[x + yi + 3i]
⇒ arg[(x – 1) + yi] = arg[x + (y + 3)i]
⇒ `tan^-1 y/(x - 1) = tan^-1 (y + 3)/x`
⇒ `y/(x - 1) = (y + 3)/x`
⇒ xy = (x – 1)(y + 3)
⇒ xy = xy + 3x – y – 3
⇒ 3x – y = 3
⇒ 3x – 3 = y
⇒ 3(x – 1) = y
⇒ `((x - 1))/y = 1/3`
⇒ x – 1 : y = 1 : 3
Hence, x – 1 : y = 1 : 3.
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