How do you know what part of the semicircle it is when $argdfrac{z-2}{z+2} = dfracpi2$ [closed]












1














I'm not quite sure how to determine whether the semicircle is below or above the x-axis, because in $argdfrac{z-2}{z+2} = dfracpi2$, it lay above the x-axis with a locus of $(4-x^2)^{1/2}$.






locus







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closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 '18 at 3:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Do you mean $arg(frac{z-2}{z+2})$ ?
    – Matti P.
    Nov 20 '18 at 11:58












  • Yeah that = pi/2
    – Ben Gillham
    Nov 20 '18 at 12:00
















1














I'm not quite sure how to determine whether the semicircle is below or above the x-axis, because in $argdfrac{z-2}{z+2} = dfracpi2$, it lay above the x-axis with a locus of $(4-x^2)^{1/2}$.






locus







share|cite|improve this question















closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 '18 at 3:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Do you mean $arg(frac{z-2}{z+2})$ ?
    – Matti P.
    Nov 20 '18 at 11:58












  • Yeah that = pi/2
    – Ben Gillham
    Nov 20 '18 at 12:00














1












1








1







I'm not quite sure how to determine whether the semicircle is below or above the x-axis, because in $argdfrac{z-2}{z+2} = dfracpi2$, it lay above the x-axis with a locus of $(4-x^2)^{1/2}$.






locus







share|cite|improve this question















I'm not quite sure how to determine whether the semicircle is below or above the x-axis, because in $argdfrac{z-2}{z+2} = dfracpi2$, it lay above the x-axis with a locus of $(4-x^2)^{1/2}$.






locus




locus






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share|cite|improve this question













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share|cite|improve this question








edited Nov 21 '18 at 1:26









user10354138

7,4142824




7,4142824










asked Nov 20 '18 at 11:46









Ben Gillham

61




61




closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 '18 at 3:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 '18 at 3:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Do you mean $arg(frac{z-2}{z+2})$ ?
    – Matti P.
    Nov 20 '18 at 11:58












  • Yeah that = pi/2
    – Ben Gillham
    Nov 20 '18 at 12:00


















  • Do you mean $arg(frac{z-2}{z+2})$ ?
    – Matti P.
    Nov 20 '18 at 11:58












  • Yeah that = pi/2
    – Ben Gillham
    Nov 20 '18 at 12:00
















Do you mean $arg(frac{z-2}{z+2})$ ?
– Matti P.
Nov 20 '18 at 11:58






Do you mean $arg(frac{z-2}{z+2})$ ?
– Matti P.
Nov 20 '18 at 11:58














Yeah that = pi/2
– Ben Gillham
Nov 20 '18 at 12:00




Yeah that = pi/2
– Ben Gillham
Nov 20 '18 at 12:00










1 Answer
1






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2














$$dfrac{x-2-iy}{x+2+iy}=dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$



Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0iff xy<0$$






share|cite|improve this answer





















  • THANKS SO MUCH!!!
    – Ben Gillham
    Nov 20 '18 at 12:05


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














$$dfrac{x-2-iy}{x+2+iy}=dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$



Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0iff xy<0$$






share|cite|improve this answer





















  • THANKS SO MUCH!!!
    – Ben Gillham
    Nov 20 '18 at 12:05
















2














$$dfrac{x-2-iy}{x+2+iy}=dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$



Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0iff xy<0$$






share|cite|improve this answer





















  • THANKS SO MUCH!!!
    – Ben Gillham
    Nov 20 '18 at 12:05














2












2








2






$$dfrac{x-2-iy}{x+2+iy}=dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$



Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0iff xy<0$$






share|cite|improve this answer












$$dfrac{x-2-iy}{x+2+iy}=dfrac{(x-iy)^2-4}{(x+2)^2+y^2}=dfrac{x^2-y^2-4-2xyi}{(x+2)^2+y^2}$$



Using atan2, $$x^2-y^2-4=0$$ and $$-2xy>0iff xy<0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 '18 at 11:59









lab bhattacharjee

223k15156274




223k15156274












  • THANKS SO MUCH!!!
    – Ben Gillham
    Nov 20 '18 at 12:05


















  • THANKS SO MUCH!!!
    – Ben Gillham
    Nov 20 '18 at 12:05
















THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 at 12:05




THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 at 12:05



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