Find the locus in the complex plain such that arg ((z/(z + 2))) = (Π/2) please help.


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Question Number 6044 by sanusihammed last updated on 10/Jun/16

$F i n d t h e l o c u s i n t h e c o m p l e x p l a i n s u c h t h a t$ $a r g (\frac{z}{z + 2}) = \frac{Π}{2}$ $p l e a s e h e l p .$
Commented by Yozzii last updated on 10/Jun/16

$a r g (z - 0) - a r g (z - (- 2)) = \frac{π}{2}$ $T h e l o c u s o f z i s a s e m i c i r c l e$ $s i n c e t h e a n g l e b e t w e e n t h e l i n e s$ $r e p r e s e n t e d b y t h e c o m p l e x n u m b e r s z - 0 a n d z - (- 2) i s$ $a l w a y s \frac{π}{2} . S o z l i e s o n t h e c i r c l e$ $∣ z - 0.5 (0 - (- 2) ∣=∣ z + 1 ∣= 0.5 (0 - (- 2)) = 1 f o r I m (z) > 0$ $o r z \in {z \in C : ∣ z + 1 ∣= 1 & I m (z) > 0} .$ $S i n c e π / 2 > 0 i t m u s t b e t h a t a r g (z) > a r g (z + 2)$ $a n d w e c a n n o t c h o o s e t h e s e m i c i r c l e ∣ z + 1 ∣= 1$ $w h e r e I m (z) < 0 .$ $B e l o w i s a n i m a g e s h o w i n g t h e l i n e$ $s e g m e n t s f o r z f r o m t h e o r i g i n a n d$ $z + 2 f r o m t h e p o i n t (- 2, 0) i n$ $t h e c o m p l e x p l a n e, r e s p e c t i v e l y .$
Commented by Yozzii last updated on 10/Jun/16

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