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

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 $(4-x^2)^{1/2}$.

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 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 at 11:58

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

add a comment |

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 $(4-x^2)^{1/2}$.

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 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 at 11:58

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

add a comment |

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 $(4-x^2)^{1/2}$.

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

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

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

edited Nov 21 at 1:26

user10354138

7,3742824

edited Nov 21 at 1:26

user10354138

7,3742824

edited Nov 21 at 1:26

user10354138

7,3742824

asked Nov 20 at 11:46

Ben Gillham

asked Nov 20 at 11:46

Ben Gillham

asked Nov 20 at 11:46

Ben Gillham

closed as off-topic by amWhy, José Carlos Santos, user10354138, Cesareo, darij grinberg Nov 21 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 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 at 11:58

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

add a comment |

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

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

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

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

add a comment |

1 Answer
1

active

oldest

votes

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

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

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

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

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

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

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

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

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

add a comment |

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

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

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

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

add a comment |

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

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

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

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

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

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

answered Nov 20 at 11:59

lab bhattacharjee

223k15156274

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

add a comment |

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

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

add a comment |

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