show-that-if-arg-z-1-z-2-z-1-z-2-pi-2-then-z-1-z-2-

Question Number 26374 by tawa tawa last updated on 24/Dec/17

show that if \arg (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2} then ∣ z_{1} ∣ = ∣ z_{2} ∣

Answered by Rasheed.Sindhi last updated on 31/Dec/17

\arg (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}

\tan^{- 1} \frac{Im (\frac{z_{1} + z_{2}}{z_{1} - z_{2}})}{Re (\frac{z_{1} + z_{2}}{z_{1} - z_{2}})} = \frac{π}{2}

\frac{Im (\frac{z_{1} + z_{2}}{z_{1} - z_{2}})}{Re (\frac{z_{1} + z_{2}}{z_{1} - z_{2}})} = \tan (\frac{π}{2}) = \infty

∴ Re (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = 0

Let z_{1} = x_{1} + {iy}_{1} & z_{2} = x_{2} + {iy}_{2}

Re (\frac{(x_{1} + {iy}_{1}) + (x_{2} + {iy}_{2})}{(x_{1} + {iy}_{1}) - (x_{2} + {iy}_{2})})

Re (\frac{(x_{1} + x_{2}) + i (y_{1} + y_{2})}{(x_{1} - x_{2}) + i (y_{1} - y_{2})})

Let x_{1} + x_{2} = u_{1}, y_{1} + y_{2} = v_{1}

x_{1} - x_{2} = u_{2} and y_{1} - y_{2} = v_{2}

Re (\frac{u_{1} + {iv}_{1}}{u_{2} + {iv}_{2}}) = Re (\frac{u_{1} + {iv}_{1}}{u_{2} + {iv}_{2}} \times \frac{u_{2} - {iv}_{2}}{u_{2} - {iv}_{2}})

= Re (\frac{u_{1} u_{2} + v_{1} v_{2} + i (u_{2} v_{1} - u_{1} v_{2})}{u_{2}^{2} + v_{2}^{2}})

= \frac{u_{1} u_{2} + v_{1} v_{2}}{u_{2}^{2} + v_{2}^{2}} = 0

u_{1} u_{2} + v_{1} v_{2} = 0

(x_{1} + x_{2}) (x_{1} - x_{2}) + (y_{1} + y_{2}) (y_{1} - y_{2}) = 0

(x_{1}^{2} - x_{2}^{2}) + (y_{1}^{2} - y_{2}^{2}) = 0

(x_{1}^{2} + y_{1}^{2}) - (x_{2}^{2} + y_{2}^{2}) = 0

(x_{1}^{2} + y_{1}^{2}) = (x_{2}^{2} + y_{2}^{2})

∣ z_{1} ∣^{2} =∣ z_{2} ∣^{2}

∣ z_{1} ∣=∣ z_{2} ∣