Find-the-locus-in-the-complex-plain-such-that-arg-z-z-2-2-please-help-

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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