de Moivre's Theorem

de Moivre's Theorem

12th Grade - University

10 Qs

quiz-placeholder

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de Moivre's Theorem

de Moivre's Theorem

Assessment

Quiz

Mathematics

12th Grade - University

Practice Problem

Medium

CCSS
HSN.CN.B.4, HSF.TF.C.9

Standards-aligned

Created by

Oxana OLAROU

Used 65+ times

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The value of 

 (1+i3)6\left(1+i\sqrt{3}\right)^6  is

64

-64

1

2

Tags

CCSS.HSN.CN.B.4

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The value of

 1(cosθ+isinθ)6\frac{1}{\left(\cos\theta+i\sin\theta\right)^6}  is

 cos6θisin6θ-\cos6\theta-i\sin6\theta  

 cos6θ+isin6θ\cos6\theta+i\sin6\theta  

 cos6θisin6θ\cos6\theta-i\sin6\theta  

 cos6θ+isin6θ-\cos6\theta+i\sin6\theta  

Tags

CCSS.HSN.CN.B.4

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (cos π6  isin π6)12 is equal to\left(\cos\ \frac{\pi}{6}\ -\ i\sin\ \frac{\pi}{6}\right)^{12}\ is\ equal\ to  


-i

-1

i

1

Tags

CCSS.HSN.CN.B.4

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

 cos4θ\cos4\theta  is eqauivalent to

 8cos4θ+8cos2θ+18\cos^4\theta+8\cos^2\theta+1  

 8cos4θ8cos2θ+18\cos^4\theta-8\cos^2\theta+1  

 8cos4θ8cos2θ18\cos^4\theta-8\cos^2\theta-1  

 8cos4θ+8cos2θ18\cos^4\theta+8\cos^2\theta-1  

Tags

CCSS.HSF.TF.C.9

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

 sin4θ\sin4\theta   is equivalent to

 8cos3θsinθ+8cosθsin3θ8\cos^3\theta\sin\theta+8\cos\theta\sin^3\theta  

 4cos3θsinθ+4cosθsin3θ4\cos^3\theta\sin\theta+4\cos\theta\sin^3\theta  

 4cos3θsinθ4cosθsin3θ4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta  

 8cos3θsinθ4cosθsin3θ8\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta  

Tags

CCSS.HSF.TF.C.9

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 If z = cosθ +isinθ, the value of z3+1z3 isIf\ z\ =\ \cos\theta\ +i\sin\theta,\ the\ value\ of\ z^3+\frac{1}{z^3}\ is  


 2cos3θ2\cos3\theta  

 cos3θ\cos3\theta  

 isin3θi\sin3\theta  

 2isin3θ2i\sin3\theta  

Tags

CCSS.HSN.CN.B.4

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 43+4i-4\sqrt{3}+4i  can be expressed in the form  reiθre^{i\theta}  as


 8e5πi68e^{\frac{5\pi i}{6}}  

 8eπi68e^{-\frac{\pi i}{6}}  

 4e5πi64e^{\frac{5\pi i}{6}}  

 4eπi64e^{-\frac{\pi i}{6}}  

Tags

CCSS.HSN.CN.B.4

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