
Concept explainers
a.
The roots in the trigonometric form by using the graph of the roots of the
a.

Answer to Problem 147RE
z0=4(cos600+isin600),z1=4(cos1500+isin1500),
z2=4(cos2400+isin2400) and z3=4(cos3300+isin3300) .
Explanation of Solution
Given information:
Write each of the roots in trigonometric form.
Calculation:
From the given information, write each of the roots in trigonometric form using the graph of roots are as shown below:
Let’s note the roots:
zi=r1(cosθi+isinθi) , where
i=0,1,2 .
From the given graph, there are three roots, then they are:
3600n=36003=1200
Starting from the x− axis going counter clockwise, the roots are:
Now, write the roots:
z0=2(cos300+isin300)z1=2(cos1500+isin1500)z2=2(cos2700+isin2700)
Thus, the roots in trigonometric forms are z0=2(cos300+isin300),z1=2(cos1500+isin1500), and z2=2(cos2700+isin2700) .
b.
The complex number by using roots. Verify the results by using graphing utility.
b.

Answer to Problem 147RE
8i .
Explanation of Solution
Given information:
Identify the complex number whose roots are given.Use a graphing utility to verify your results.
Calculation:
From the given information, identify the complex number whose roots are given and by using graphing utility verify the result are as shown below:
From the part (a) found the roots:
z0=2(cos300+isin300)z1=2(cos1500+isin1500)z2=2(cos2700+isin2700)
These equations verify the equation:
z3=ω
Now find
ω , by using the fact that any
zi verifies the above equation:
ω=z40=z41=z42=z43
To find ω ,
ω=z30=[2(cos300+isin300)3]z30=23[cos(3⋅300)+isin(3⋅300)]z30=8(cos900+isin900)z30=8(0+i)z30=z=8i
Thus , the complex number is 8i .
Chapter 6 Solutions
EBK PRECALCULUS W/LIMITS
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