The number z = −2 + 2i has modulus |z| = 2√2 and argument arg z =(5/4)π, now I dont understand how they calculate the argument, I get that -2+2i is in the second quadrant and the formula for that is arctan(b/a)+π. now we have arctan(-1) first. and when i look at the tan unit circle I get (3/4)π, I don't understand why this is wrong,
The number z = −2 + 2i has modulus |z| = 2√2 and argument arg z =(5/4)π, now I dont understand how they calculate the argument, I get that -2+2i is in the second quadrant and the formula for that is arctan(b/a)+π. now we have arctan(-1) first. and when i look at the tan unit circle I get (3/4)π, I don't understand why this is wrong,
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The number z = −2 + 2i has modulus |z| = 2√2 and argument arg z =(5/4)π, now I dont understand how they calculate the argument, I get that -2+2i is in the second quadrant and the formula for that is arctan(b/a)+π.
now we have arctan(-1) first. and when i look at the tan unit circle I get (3/4)π, I don't understand why this is wrong,
Transcribed Image Text:C
ო|ო
√3
3
0 π
ო|ო
√3
5π
6
7π
-√3 d
2π
3π 3
6 5T
4
√3
not
defined
-135
-210
-225
150
180°
4π
3
120
-240
TU
2
90°
09
300
73
not
defined
π
45°
30
330
315
√3
74
11π
7 6
270 54
3π 3
2
76
-√3
-1
33
O
-√3
3
Unit Circle With Tangent - Values, Chart, Calculator
Transcribed Image Text:arg (z) = π
arg (z)
2
= arctan
arg (z) = arctan
3
210
π
arg (2) = 7
+ π
(1) - 7
arg (z)
arg (z)
arg (z) = -7
2
=
1
arctan|
4
= arctan
()
arg (z) = 0
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