
Trigonometrijski oblik kompleksnog brojasni
Authored by Rijana Kosovic
Mathematics
9th Grade
Used 1+ times

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7 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koji je trigonometrijski oblik kompleksnog broja z = 3 + 4i?
5(cos(arctan(3/4)) + i*sin(arctan(3/4)))
5(cos(arctan(4/3)) - i*sin(arctan(4/3)))
5(cos(arctan(3/4)) - i*sin(arctan(3/4)))
5(cos(arctan(4/3)) + i*sin(arctan(4/3)))
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koji je trigonometrijski oblik kompleksnog broja z = -2 - 2i?
z = 2√2(cos(π/4) + i sin(π/4))
z = -2√2(cos(π/4) + i sin(π/4))
z = -2√2(cos(5π/4) + i sin(5π/4))
z = 2√2(cos(3π/4) + i sin(3π/4))
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koji je korijen kompleksnog broja z = -1 - i?
sqrt(sqrt(2)) * (cos(pi/8) + i*sin(pi/8))
i - 1
1 - i
-1 + i
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koji je korijen kompleksnog broja z = 2(cos(π/6) + i sin(π/6))?
sqrt(2)(cos(π/12) + i sin(π/12))
2(cos(π/3) + i sin(π/3))
2(cos(π/4) + i sin(π/4))
2(cos(π/6) + i sin(π/6))
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koji je korijen kompleksnog broja z = 3(cos(π/3) + i sin(π/3))?
3(cos(π/3) + i sin(π/3))
3(cos(π/3) + i sin(π/4))
3(cos(π/3) + i sin(π/6))
3(cos(π/3) + i sin(π/2))
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koja je Moavreova formula za potenciranje kompleksnih brojeva?
(cos(theta) - i*sin(theta))^n = cos(n*theta) + i*sin(n*theta)
(cos(theta) + i*sin(theta))^n = cos(n*theta) + i*sin(n*theta)
(cos(theta) + i*sin(theta))^n = cos(n*theta) - i*sin(n*theta)
(cos(theta) - i*sin(theta))^n = cos(n*theta) - i*sin(n*theta)
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Koja je Moavreova formula za korijenovanje kompleksnih brojeva?
z^n = r^n * (cos(n*theta) + i*sin(n*theta))
z^n = r^n * (cos(n*theta) + i*sin(n*theta))^4
z^n = r^n * (cos(n*theta) + i*sin(n*theta))^3
z^n = r^n * (cos(n*theta) + i*sin(n*theta))^2
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