LOG Antonic Borisavljevic 4 6
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Transcript of LOG Antonic Borisavljevic 4 6
8/15/2019 LOG Antonic Borisavljevic 4 6
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|z − z1|{=, <, ≤, >, ≥}
|z − z1| = R
|z − z1| = R|x + iy − x1 − iy1| = R
(x − x1)2 + (y − y1)2 = R/2
(x − x1)2 + (y − y1)2 = R2
|z − z1|<R|x + iy − x1 − iy1| < R
(x − x1)2 + (y − y1)2 < R/2
(x − x1)2 + (y − y1)2 < R2
|z − z1|≤R|x + iy − x1 − iy1|≤R
(x − x1)2 + (y − y1)2≤R
2
(x − x1)2 + (y − y1)2 ≤ R2
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|z − z1|>R|x + iy − x1 − iy1| > R
(x −
x1)2 + (y −
y1)2 > R/2
(x − x1)2 + (y − y1)2 > R2
|z − z1| ≥ R|x + iy − x1 − iy1| ≥ R
(x − x1)2 + (y − y1)2 ≥ R/2
(x − x1)2 + (y − y1)2 ≥ R2
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|z − 1| ≤ 1, |z − i| ≥ 1 |x + iy + i| ≥
1o |z − 1| ≤ 1|x + iy − 1|≤ 1
(x − 1)2 + y2 ≤ 1/2
(x − 1)2 + y2 ≤ 12o |z − i| ≥ 1|x + iy − i| ≥ 1
x2 + (y − 1)2 ≥ 1/2
x2 + (y − 1)2 ≥ 1
1o2 < |z| ≤ 33 ≥ |z| > 23 ≥ |x + iy| > 2
3 ≥
x2 + y2 > 2/2
9 ≥ x2 + y2 > 42o
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−π
3 ≤ arg(z) ≤ π
4
1o0 ≤ arg(z) ≤ π
4
2o |z − 1 − i| ≤ 1|x + iy − 1 − i| ≤ 1
(x − 1)2 + (y − 1)2 ≤ 1/2
(x − 1)2 + (y − 1)2 ≤ 13oRe(z − 1) < 0
1o |z − 1 − i| ≤ 2|x + iy − 1 − i| ≤ 2
(x − 1)2 + (y − 1)2 ≤ 2/2
(x − 1)2 (y − 1)2≤ 42o |z + 3 + 3i| > 3|x + iy + 3 + 3i|> 3
2
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(x + 3)2 + (y + 3)2 > 9
1o 3π
4 ≤ arg (z) < 5π
3
2o|z + 2 − i|≤ 4
2
(x + 2)2 + (y − 1)2≤ 16
3oRe(z + 3 − 4i) > 0x + 3 > 0x > −3
|z − z1| = |z − z2|
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1o|z − 2 + i| |z − 4 + 5i||x + iy − 2 + i| |x + iy − 4 + 5i|
(x −
2)2 + (Y + 1)2
(x
− 4)2 + (y + 52
2
x2 − 4x + 4 + y2 + 2y + 1 = x2 − 8x + 16 + y2 + 10y + 254x − 8y − 36 = 0x − 2y − 9 = 02o|z − 1 + 4i| |z − 5 + 2i||x + iy − 1 + 4i| |x + iy − 5 + 2i|
(x − 1)2 + (y + 4)2
(x − 5)2 + (y + 2)2
2
x2 − 2x + 1 + y2 + 8y + 16 = x2 − 10x + 25 + y2 + 4y + 48x + 4y − 12 = 02x + y − 3 = 01oi2o
x − 2y − 9 = 0 => x = 2y + 92 ∗ (2y + 9) + y − 3 = 0
4y + 18 + y − 3 = 05y + 15 = 05y = −15y = −3x = 3z = 3 − 3i
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1o |z − 4i| ≥ |z − 6 − 2i|x2 + (y − 4)2 ≥ (x − 6)2 + (y − 2)2
x2+y2 − 8y + 16 ≥ x2 − 12x + 36 + y2 − 4y + 4−8y + 16 ≥ −12x − 4y + 40−12x − 4y ≥ 243x − y ≥ 620|z − 3 + 2i|≤ √
5 (x − 3)2 + (y + 2)2 ≤ √
5/2
(x − 3)2 + (y + 2)2 ≤ 53oRe(z − 4) < 0x < 4
z = r(cos α + i sin α)
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x = r cos αy = r sin αr =
x2 + y2
cos α = x
r
sin α = y
r
z = 5, x = 5, y = 0z = r(cos α + i sin α)
r =
x2 + y2
r = 5cos α = x
r= 5
5 = 1
sin α = y
r= 0
z = 5cis 0
z = 2i, x = 0, y = 2z = r(cos α + i sin α)
r =
x2 + y2
r = 2cos α = x
r= 0
sin α = y
r= 1
z = 2 cis π2
z = −7, x = −7, y = 0z = r(cos α + i sin α)
r =
x2 + y2
r = 7cos α = x
r= −1
sin α = y
r= 0
z = 7 cis π
z = −3i, x = 0, y = −3z = r(cos α + i sin α)
r =
x2 + y2
r = 3
cos α = xr = 0
sin α = y
r= −1
z = 3 cis 3π2
z = 1 + i, x = 1, y = 1z = r(cos α + i sin α)
r =
x2 + y2
r =√
2cos α = x
r=
√ 2
2
sin α = y
r=
√ 2
2
z =√
2cis π4
z = √ 2 − √ 6i, x = √ 2, y = −√ 6z = r(cos α + i sin α)
r =
x2 + y2
r = 2√
2cos α = x
r= 1
2
sin α = y
r=
√ 3
2
z = 2√
2cis π3
z = −√ 57 − √
17i, x = −√ 57, y = −√
17z = r(cos α + i sin α)
r =
x2 + y2
r = 2√
17cos α = x
r
= −
√ 3
2