PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
Chapter 6.6, Problem 50E
Solution
To find: The fifth roots of the given
The fifth roots of the given complex number are 5√2(cosπ15+isinπ15) , 5√2(cos7π15+isin7π15) , 5√2(cos13π15+isin13π15) , 5√2(cos19π15+isin19π15) and (1−√3i)5√16 .
Given information:
The complex number is 1+√3i .
Formula used:
The polar form of a complex number z=a+bi is given by,
z=r(cosθ+isinθ)
Here, r=√a2+b2 and tanθ=ba .
The nth roots of a complex number z=r(cosθ+isinθ) which are n distinct complex numbers are given by,
n√r(cosθ+2πkn+isinθ+2πkn)
Here, k=0, 1, 2,..., n−1
Calculation:
From the given complex number,
a=1b=√3
Substitute 1 for a and √3 for b in equation r=√a2+b2 to find modulus r .
r=√(1)2+(√3)2=√4=2
Substitute 1 for a and √3 for b in equation tanθ=ba to find argument.
tanθ=√31θ=tan−1(√3)=π3
Substitute 2 for r and π3 for θ in equation z=r(cosθ+isinθ) .
z=2(cosπ3+isinπ3)
From the above complex number,
r=2θ=π3n=5
Use k=0, 1, 2, 3 and 4 to determine the cube roots.
Substitute 2 for r , π3 for θ , 5 for n , and 0 for k in expression n√r(cosθ+2πkn+isinθ+2πkn) to find first root.
z1=5√2(cosπ3+2π⋅05+isinπ3+2π⋅05)=5√2(cosπ15+isinπ15)
Substitute 2 for r , π3 for θ , 5 for n , and 1 for k in expression n√r(cosθ+2πkn+isinθ+2πkn) to find second root.
z2=5√2(cosπ3+2π⋅15+isinπ3+2π⋅15)=5√2(cos7π15+isin7π15)
Substitute 2 for r , π3 for θ , 5 for n , and 2 for k in expression n√r(cosθ+2πkn+isinθ+2πkn) to find third root.
z3=5√2(cosπ3+2π⋅25+isinπ3+2π⋅25)=5√2(cos13π15+isin13π15)
Substitute 2 for r , π3 for θ , 5 for n , and 3 for k in expression n√r(cosθ+2πkn+isinθ+2πkn) to find fourth root.
z4=5√2(cosπ3+2π⋅35+isinπ3+2π⋅35)=5√2(cos19π15+isin19π15)
Substitute 2 for r , π3 for θ , 5 for n , and 4 for k in expression n√r(cosθ+2πkn+isinθ+2πkn) to find fifth root.
z5=5√2(cosπ3+2π⋅45+isinπ3+2π⋅45)=5√2(cos5π3+isin5π3)=(1−√3i)5√16
Therefore, the fifth roots of the given complex number are 5√2(cosπ15+isinπ15) , 5√2(cos7π15+isin7π15) , 5√2(cos13π15+isin13π15) , 5√2(cos19π15+isin19π15) and (1−√3i)5√16 .
Chapter 6 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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