Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question

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,

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
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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
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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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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