Complex locus with sum of arguments
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Find the locus of $z$ when: $$arg(z+2) + arg(z-2) = pi. $$
I tried substituting $ z = x + iy$ but the algebra is fairly messy and I'm not sure how to work through it. Is there a better approach to solving the question.
Thanks.
complex-numbers complex-geometry
edited Nov 26 at 9:06
Tianlalu
2,9801936
asked Nov 26 at 9:03
ultralight
396
add a comment |
up vote
0
down vote
favorite
Find the locus of $z$ when: $$arg(z+2) + arg(z-2) = pi. $$
I tried substituting $ z = x + iy$ but the algebra is fairly messy and I'm not sure how to work through it. Is there a better approach to solving the question.
Thanks.
complex-numbers complex-geometry
edited Nov 26 at 9:06
Tianlalu
2,9801936
asked Nov 26 at 9:03
ultralight
396
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the locus of $z$ when: $$arg(z+2) + arg(z-2) = pi. $$
I tried substituting $ z = x + iy$ but the algebra is fairly messy and I'm not sure how to work through it. Is there a better approach to solving the question.
Thanks.
complex-numbers complex-geometry
edited Nov 26 at 9:06
Tianlalu
2,9801936
asked Nov 26 at 9:03
ultralight
396
Find the locus of $z$ when: $$arg(z+2) + arg(z-2) = pi. $$
I tried substituting $ z = x + iy$ but the algebra is fairly messy and I'm not sure how to work through it. Is there a better approach to solving the question.
Thanks.
complex-numbers complex-geometry
complex-numbers complex-geometry
edited Nov 26 at 9:06
Tianlalu
2,9801936
asked Nov 26 at 9:03
ultralight
396
edited Nov 26 at 9:06
Tianlalu
2,9801936
asked Nov 26 at 9:03
ultralight
396
edited Nov 26 at 9:06
Tianlalu
2,9801936
edited Nov 26 at 9:06
Tianlalu
2,9801936
edited Nov 26 at 9:06
Tianlalu
2,9801936
2,9801936
asked Nov 26 at 9:03
ultralight
396
asked Nov 26 at 9:03
ultralight
396
asked Nov 26 at 9:03
ultralight
396
396
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
From $$arg (z+2) + arg (z-2) = arg (z^2-4) = pi $$
we deduce that $;z^2-4;$ is a negative real number. This occurs when
$zin mathbb{R},; -2<z<2,$ or
$z=ialpha,; alpha in mathbb{R},$ is pure imaginary.
answered Nov 26 at 12:54
user376343
2,6412819
add a comment |
up vote
0
down vote
HINT
A trivial case is when $z$ is real.
Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and
$$arg(z+2) + arg(z-2) = pi iff arctanleft(frac{y}{x+2}right)+arctanleft(frac{y}{x-2}right)+pi=pi$$
answered Nov 26 at 10:05
gimusi
91k74495
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
From $$arg (z+2) + arg (z-2) = arg (z^2-4) = pi $$
we deduce that $;z^2-4;$ is a negative real number. This occurs when
$zin mathbb{R},; -2<z<2,$ or
$z=ialpha,; alpha in mathbb{R},$ is pure imaginary.
answered Nov 26 at 12:54
user376343
2,6412819
add a comment |
up vote
1
down vote
accepted
From $$arg (z+2) + arg (z-2) = arg (z^2-4) = pi $$
we deduce that $;z^2-4;$ is a negative real number. This occurs when
$zin mathbb{R},; -2<z<2,$ or
$z=ialpha,; alpha in mathbb{R},$ is pure imaginary.
answered Nov 26 at 12:54
user376343
2,6412819
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
From $$arg (z+2) + arg (z-2) = arg (z^2-4) = pi $$
we deduce that $;z^2-4;$ is a negative real number. This occurs when
$zin mathbb{R},; -2<z<2,$ or
$z=ialpha,; alpha in mathbb{R},$ is pure imaginary.
answered Nov 26 at 12:54
user376343
2,6412819
From $$arg (z+2) + arg (z-2) = arg (z^2-4) = pi $$
we deduce that $;z^2-4;$ is a negative real number. This occurs when
$zin mathbb{R},; -2<z<2,$ or
$z=ialpha,; alpha in mathbb{R},$ is pure imaginary.
answered Nov 26 at 12:54
user376343
2,6412819
edited Nov 26 at 20:10
answered Nov 26 at 12:54
user376343
2,6412819
answered Nov 26 at 12:54
user376343
2,6412819
answered Nov 26 at 12:54
user376343
2,6412819
2,6412819
add a comment |
add a comment |
up vote
0
down vote
HINT
A trivial case is when $z$ is real.
Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and
$$arg(z+2) + arg(z-2) = pi iff arctanleft(frac{y}{x+2}right)+arctanleft(frac{y}{x-2}right)+pi=pi$$
answered Nov 26 at 10:05
gimusi
91k74495
add a comment |
up vote
0
down vote
HINT
A trivial case is when $z$ is real.
Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and
$$arg(z+2) + arg(z-2) = pi iff arctanleft(frac{y}{x+2}right)+arctanleft(frac{y}{x-2}right)+pi=pi$$
answered Nov 26 at 10:05
gimusi
91k74495
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
A trivial case is when $z$ is real.
Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and
$$arg(z+2) + arg(z-2) = pi iff arctanleft(frac{y}{x+2}right)+arctanleft(frac{y}{x-2}right)+pi=pi$$
answered Nov 26 at 10:05
gimusi
91k74495
HINT
A trivial case is when $z$ is real.
Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and
$$arg(z+2) + arg(z-2) = pi iff arctanleft(frac{y}{x+2}right)+arctanleft(frac{y}{x-2}right)+pi=pi$$
answered Nov 26 at 10:05
gimusi
91k74495
answered Nov 26 at 10:05
gimusi
91k74495
answered Nov 26 at 10:05
gimusi
91k74495
answered Nov 26 at 10:05
gimusi
91k74495
91k74495
add a comment |
add a comment |
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