a Answer The fifth roots of 128 ( − 1 + i ) are 2 2 ( cos 3 π 20 + i sin 3 π 20 ) , 2 2 ( cos 11 π 20 + i sin 11 π 20 ) , 2 2 ( cos 19 π 20 + i sin 19 π 20 ) , 2 2 ( cos 27 π 20 + i sin 27 π 20 ) and 2 2 ( cos 7 π 20 + i sin 7 π 20 ) . Explanation Given information: The complex number is 128 ( − 1 + i ) Calculation: Calculate the value of r . r = | − 128 + 128 i | = ( − 128 ) 2 + ( 128 ) 2 = 128 2 Calculate the value of θ . θ = arctan ( − 1 ) = 3 π 4 The trigonometric form of z . z = 128 2 ( cos 3 π 4 + i sin 3 π 4 ) The n t h roots of a complex number z = r ( cos θ + i sin θ ) are given by z k = r n ( cos θ + 2 π k n + i sin θ + 2 π k n ) ……. (1) Here, k = 0 , 1 , 2...... n − 1 . Consider the complex number. 128 ( − 1 + i ) The fourth roots of 128 ( − 1 + i ) is to be evaluated n = 5 and r = 128 2 by the n t h roots formula the roots are z k = 128 2 4 ( cos 3 π 4 + 2 π k 5 + i sin 3 π 4 + 2 π k 5 ) ……. (2) Substitute 0 for k in equation (2). z 0 = 128 2 5 ( cos 3 π 4 + 2 π ( 0 ) 5 + i sin 3 π 4 + 2 π ( 0 ) 5 ) z 0 = 2 2 ( cos 3 π 20 + i sin 3 π 20 ) Substitute 1 for k in equation (2). z 1 = 128 2 5 ( cos 3 π 4 + 2 π ( 1 ) 5 + i sin 3 π 4 + 2 π ( 1 ) 5 ) z 1 = 2 2 ( cos 11 π 20 + i sin 11 π 20 ) Substitute 2 for k in equation (2). z 2 = 128 2 5 ( cos 3 π 4 + 2 π ( 2 ) 5 + i sin 3 π 4 + 2 π ( 2 ) 5 ) z 2 = 2 2 ( cos 19 π 20 + i sin 19 π 20 ) Substitute 3 for k in equation (2). z 3 = 128 2 5 ( cos 3 π 4 + 2 π ( 3 ) 5 + i sin 3 π 4 + 2 π ( 3 ) 5 ) z 3 = 2 2 ( cos 27 π 20 + i sin 27 π 20 ) Substitute 4 for k in equation (2). z 4 = 128 2 5 ( cos 3 π 4 + 2 π ( 4 ) 5 + i sin 3 π 4 + 2 π ( 4 ) 5 ) z 4 = 2 2 ( cos 7 π 20 + i sin 7 π 20 ) Therefore, the fifth roots of 128 ( − 1 + i ) are 2 2 ( cos 3 π 20 + i sin 3 π 20 ) , 2 2 ( cos 11 π 20 + i sin 11 π 20 ) , 2 2 ( cos 19 π 20 + i sin 19 π 20 ) , 2 2 ( cos 27 π 20 + i sin 27 π 20 ) and 2 2 ( cos 7 π 20 + i sin 7 π 20 ) . b To determine The graph for the complex roots. Answer The graph for the complex roots is shown in Figure (1). Explanation Given information: The complex number is 128 ( − 1 + i ) Calculation: The fifth roots of 128 ( − 1 + i ) are 2 2 ( cos 3 π 20 + i sin 3 π 20 ) , 2 2 ( cos 11 π 20 + i sin 11 π 20 ) , 2 2 ( cos 19 π 20 + i sin 19 π 20 ) , 2 2 ( cos 27 π 20 + i sin 27 π 20 ) and 2 2 ( cos 7 π 20 + i sin 7 π 20 ) . Draw the graph for the roots. Figure-(1) Therefore, the graph for the complex roots is shown in Figure (1). c To determine The standard form of the complex roots. Answer The standard forms of the roots are 2.5201 + 1.2841 i , − 0.4425 + 2.7936 i , − 2.7936 − 0.4425 i , − 1.2841 − 2.5201 i and 2 − 2 i . Explanation Given information: The fifth roots of 128 ( − 1 + i ) are 2 2 ( cos 3 π 20 + i sin 3 π 20 ) , 2 2 ( cos 11 π 20 + i sin 11 π 20 ) , 2 2 ( cos 19 π 20 + i sin 19 π 20 ) , 2 2 ( cos 27 π 20 + i sin 27 π 20 ) and 2 2 ( cos 7 π 20 + i sin 7 π 20 ) . Calculation: Calculate the standard form of the root z 0 . z 0 = 2 2 ( cos 3 π 20 + i sin 3 π 20 ) = 2.5201 + 1.2841 i Calculate the standard form of the root z 1 . z 1 = 2 2 ( cos 11 π 20 + i sin 11 π 20 ) = − 0.4425 + 2.7936 i Calculate the standard form of the root z 2 . z 2 = 2 2 ( cos 19 π 20 + i sin 19 π 20 ) = − 2.7936 − 0.4425 i Calculate the standard form of the root z 3 . z 3 = 2 2 ( cos 27 π 20 + i sin 27 π 20 ) = − 1.2841 − 2.5201 i Calculate the standard form of the root z 4 . z 4 = 2 2 ( cos 7 π 20 + i sin 7 π 20 ) = 2 − 2 i Therefore, the standard forms of the roots are 2.5201 + 1.2841 i , − 0.4425 + 2.7936 i , − 2.7936 − 0.4425 i , − 1.2841 − 2.5201 i and 2 − 2 i .

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Question

Chapter 6.5, Problem 151E

To determine

The fifth roots of the complex number.

Expert Solution

Answer to Problem 151E

The fifth roots of $128(−1+i)$ are $22(cos3π20+isin3π20)$ , $22(cos11π20+isin11π20)$ , $22(cos19π20+isin19π20)$ , $22(cos27π20+isin27π20)$ and $22(cos7π20+isin7π20)$ .

Explanation of Solution

Given information:

The complex number is $128(−1+i)$

Calculation:

Calculate the value of $r$ .

$r=|−128+128i|=(−128)2+(128)2=1282$

Calculate the value of $θ$ .

$θ=arctan(−1)=3π4$

The trigonometric form of $z$ .

$z=1282(cos3π4+isin3π4)$

The $nth$ roots of a complex number $z=r(cosθ+isinθ)$ are given by

$zk=rn(cosθ+2πkn+isinθ+2πkn)$ ……. (1)

Here, $k=0,1,2......n−1$ .

Consider the complex number.

$128(−1+i)$

The fourth roots of $128(−1+i)$ is to be evaluated $n=5$ and $r=1282$ by the $nth$ roots formula the roots are

$zk=12824(cos3π4+2πk5+isin3π4+2πk5)$ ……. (2)

Substitute $0$ for $k$ in equation (2).

$z0=12825(cos3π4+2π(0)5+isin3π4+2π(0)5)z0=22(cos3π20+isin3π20)$

Substitute $1$ for $k$ in equation (2).

$z1=12825(cos3π4+2π(1)5+isin3π4+2π(1)5)z1=22(cos11π20+isin11π20)$

Substitute $2$ for $k$ in equation (2).

$z2=12825(cos3π4+2π(2)5+isin3π4+2π(2)5)z2=22(cos19π20+isin19π20)$

Substitute $3$ for $k$ in equation (2).

$z3=12825(cos3π4+2π(3)5+isin3π4+2π(3)5)z3=22(cos27π20+isin27π20)$

Substitute $4$ for $k$ in equation (2).

$z4=12825(cos3π4+2π(4)5+isin3π4+2π(4)5)z4=22(cos7π20+isin7π20)$

Therefore, the fifth roots of $128(−1+i)$ are $22(cos3π20+isin3π20)$ , $22(cos11π20+isin11π20)$ , $22(cos19π20+isin19π20)$ , $22(cos27π20+isin27π20)$ and $22(cos7π20+isin7π20)$ .

To determine

The graph for the complex roots.

Expert Solution

Answer to Problem 151E

The graph for the complex roots is shown in Figure (1).

Explanation of Solution

Given information:

The complex number is $128(−1+i)$

Calculation:

The fifth roots of $128(−1+i)$ are $22(cos3π20+isin3π20)$ , $22(cos11π20+isin11π20)$ , $22(cos19π20+isin19π20)$ , $22(cos27π20+isin27π20)$ and $22(cos7π20+isin7π20)$ .

Draw the graph for the roots.

Precalculus with Limits: A Graphing Approach, Chapter 6.5, Problem 151E

Figure-(1)

Therefore, the graph for the complex roots is shown in Figure (1).

To determine

The standard form of the complex roots.

Expert Solution

Answer to Problem 151E

The standard forms of the roots are $2.5201+1.2841i$ , $−0.4425+2.7936i$ , $−2.7936−0.4425i$ , $−1.2841−2.5201i$ and $2−2i$ .

Explanation of Solution

Given information:

The fifth roots of $128(−1+i)$ are $22(cos3π20+isin3π20)$ , $22(cos11π20+isin11π20)$ , $22(cos19π20+isin19π20)$ , $22(cos27π20+isin27π20)$ and $22(cos7π20+isin7π20)$ .

Calculation:

Calculate the standard form of the root $z0$ .

$z0=22(cos3π20+isin3π20)=2.5201+1.2841i$

Calculate the standard form of the root $z1$ .

$z1=22(cos11π20+isin11π20)=−0.4425+2.7936i$

Calculate the standard form of the root $z2$ .

$z2=22(cos19π20+isin19π20)=−2.7936−0.4425i$

Calculate the standard form of the root $z3$ .

$z3=22(cos27π20+isin27π20)=−1.2841−2.5201i$

Calculate the standard form of the root $z4$ .

$z4=22(cos7π20+isin7π20)=2−2i$

Therefore, the standard forms of the roots are $2.5201+1.2841i$ , $−0.4425+2.7936i$ , $−2.7936−0.4425i$ , $−1.2841−2.5201i$ and $2−2i$ .

Chapter 6.5, Problem 150E

Chapter 6.5, Problem 152E

Chapter 6 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E

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