Question

If arg (z-1) = arg (z + 3i), find (x - 1): y where z= x + iy.
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Answer 1

Answer:

i think so.ot could be wrong

Answer 2

z = x + iy

arg(z - 1) = arg(z + 3i)

arg (x + iy - 1) = arg(x + iy + 3i)

arg((x - 1) + iy) = arg(x + (y + 3)i)

${tan}^{ - 1} ( \frac{y}{x - 1} ) = {tan}^{ - 1} ( \frac{y + 3}{x} ) \: \: \: ... \: arg(a + ib) = {tan}^{ - 1} ( \frac{b}{a} ) \\ \\ taking \: tangents \: on \: both \: sides \: we \: get \\ \\ \frac{y}{x - 1} = \frac{y + 3}{x} \\ xy = (y + 3)(x - 1) \\ xy = xy - y + 3x - 3 \\ xy - xy = 3(x - 1) - y \\ 3(x - 1) - y = 0 \\ 3(x - 1) = y \\ \frac{x - 1}{y} = \frac{1}{3}$

Hence, (x - 1) : y = 1 : 3