Complex locus with sum of arguments
$begingroup$
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 '18 at 9:06
Tianlalu
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
$endgroup$
add a comment |
$begingroup$
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 '18 at 9:06
Tianlalu
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
$endgroup$
add a comment |
$begingroup$
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 '18 at 9:06
Tianlalu
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
$endgroup$
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 '18 at 9:06
Tianlalu
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
edited Nov 26 '18 at 9:06
Tianlalu
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
edited Nov 26 '18 at 9:06
Tianlalu
3,08621038
edited Nov 26 '18 at 9:06
Tianlalu
3,08621038
edited Nov 26 '18 at 9:06
Tianlalu
3,08621038
3,08621038
asked Nov 26 '18 at 9:03
ultralightultralight
396
asked Nov 26 '18 at 9:03
ultralightultralight
396
asked Nov 26 '18 at 9:03
ultralightultralight
396
396
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 12:54
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
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 '18 at 10:05
gimusigimusi
92.9k94494
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
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 '18 at 12:54
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
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 '18 at 12:54
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
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 '18 at 12:54
user376343user376343
3,3833826
$endgroup$
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 '18 at 12:54
user376343user376343
3,3833826
edited Nov 26 '18 at 20:10
answered Nov 26 '18 at 12:54
user376343user376343
3,3833826
answered Nov 26 '18 at 12:54
user376343user376343
3,3833826
answered Nov 26 '18 at 12:54
user376343user376343
3,3833826
3,3833826
add a comment |
add a comment |
$begingroup$
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 '18 at 10:05
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
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 '18 at 10:05
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
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 '18 at 10:05
gimusigimusi
92.9k94494
$endgroup$
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 '18 at 10:05
gimusigimusi
92.9k94494
answered Nov 26 '18 at 10:05
gimusigimusi
92.9k94494
answered Nov 26 '18 at 10:05
gimusigimusi
92.9k94494
answered Nov 26 '18 at 10:05
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
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