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







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














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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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






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


















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



Popular posts from this blog

'app-layout' is not a known element: how to share Component with different Modules

Image
0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
Read more

android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
Read more

WPF add header to Image with URL pettitions [duplicate]

Image
up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
Read more