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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 $(4-x^2)^{1/2}$.
locus
edited Nov 21 '18 at 1:26
user10354138
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
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.
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 $(4-x^2)^{1/2}$.
locus
edited Nov 21 '18 at 1:26
user10354138
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
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.
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
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 $(4-x^2)^{1/2}$.
locus
edited Nov 21 '18 at 1:26
user10354138
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
61
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
edited Nov 21 '18 at 1:26
user10354138
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
61
edited Nov 21 '18 at 1:26
user10354138
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
61
edited Nov 21 '18 at 1:26
user10354138
7,4142824
edited Nov 21 '18 at 1:26
user10354138
7,4142824
edited Nov 21 '18 at 1:26
user10354138
7,4142824
7,4142824
asked Nov 20 '18 at 11:46
Ben Gillham
61
asked Nov 20 '18 at 11:46
Ben Gillham
61
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
add a comment |
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
add a comment |
1 Answer
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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 '18 at 11:59
lab bhattacharjee
223k15156274
THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 at 12:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 '18 at 11:59
lab bhattacharjee
223k15156274
THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 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 '18 at 11:59
lab bhattacharjee
223k15156274
THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 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 '18 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 '18 at 11:59
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 11:59
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 11:59
lab bhattacharjee
223k15156274
answered Nov 20 '18 at 11:59
lab bhattacharjee
223k15156274
223k15156274
THANKS SO MUCH!!!
– Ben Gillham
Nov 20 '18 at 12:05
add a comment |
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
add a comment |
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