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Trigonometrijski oblik kompleksnog brojasni

Authored by Rijana Kosovic

Mathematics

9th Grade

Used 1+ times

Trigonometrijski oblik kompleksnog brojasni
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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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