Question Number 6044 by sanusihammed last updated on 10/Jun/16
$${Find}\:{the}\:{locus}\:{in}\:{the}\:{complex}\:{plain}\:{such}\:{that}\: \\ $$$${arg}\:\left(\frac{{z}}{{z}\:+\:\mathrm{2}}\right)\:=\:\frac{\Pi}{\mathrm{2}} \\ $$$$ \\ $$$${please}\:{help}. \\ $$
Commented by Yozzii last updated on 10/Jun/16
$${arg}\left({z}−\mathrm{0}\right)−{arg}\left({z}−\left(−\mathrm{2}\right)\right)=\frac{\pi}{\mathrm{2}} \\ $$$${The}\:{locus}\:{of}\:{z}\:{is}\:{a}\:{semicircle}\: \\ $$$${since}\:{the}\:{angle}\:{between}\:{the}\:{lines} \\ $$$${represented}\:{by}\:{the}\:{complex}\:{numbers}\:{z}−\mathrm{0}\:{and}\:{z}−\left(−\mathrm{2}\right)\:{is}\: \\ $$$${always}\:\frac{\pi}{\mathrm{2}}.\:{So}\:{z}\:{lies}\:{on}\:{the}\:{circle}\: \\ $$$$\mid{z}−\mathrm{0}.\mathrm{5}\left(\mathrm{0}−\left(−\mathrm{2}\right)\mid=\mid{z}+\mathrm{1}\mid=\mathrm{0}.\mathrm{5}\left(\mathrm{0}−\left(−\mathrm{2}\right)\right)=\mathrm{1}\:{for}\:{Im}\left({z}\right)>\mathrm{0}\right. \\ $$$${or}\:{z}\in\left\{{z}\in\mathbb{C}:\:\mid{z}+\mathrm{1}\mid=\mathrm{1}\:\&\:{Im}\left({z}\right)>\mathrm{0}\right\}. \\ $$$${Since}\:\pi/\mathrm{2}>\mathrm{0}\:{it}\:{must}\:{be}\:{that}\:{arg}\left({z}\right)>{arg}\left({z}+\mathrm{2}\right) \\ $$$${and}\:{we}\:{cannot}\:{choose}\:{the}\:{semicircle}\:\mid{z}+\mathrm{1}\mid=\mathrm{1} \\ $$$${where}\:{Im}\left({z}\right)<\mathrm{0}. \\ $$$${Below}\:{is}\:{an}\:{image}\:{showing}\:{the}\:{line} \\ $$$${segments}\:{for}\:{z}\:{from}\:{the}\:{origin}\:{and}\: \\ $$$$\:{z}+\mathrm{2}\:{from}\:{the}\:{point}\:\left(−\mathrm{2},\mathrm{0}\right)\:{in} \\ $$$${the}\:{complex}\:{plane},{respectively}. \\ $$
Commented by Yozzii last updated on 10/Jun/16
