(a)
The trigonometric forms of thecomplex numbers.
(a)

Answer to Problem 46E
The trigonometric formis (√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]
Explanation of Solution
Given info:
(√3+i)(1+i)
Formula used:
The trigonometric form of the
z=r(cosθ+isinθ)
Where r=√a2+b2 and θ=tan−1(ba)
Calculation:
We have
z=(√3+i)(1+i)√3+i_a=√3,b=1r=√a2+b2=√(√3)2+12=2θ=tan−1(1√3)=π6z=r(cosθ+isinθ)√3+i=2(cosπ6+isinπ6)1+i_a=1,b=1r=√a2+b2=√12+12=√2θ=tan−1(11)=π4z=r(cosθ+isinθ)1+i=√2(cosπ4+isinπ4)z=(√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]
The trigonometric form is (√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]
Conclusion:
Thus,the trigonometric form is (√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]
(b)
The product using the trigonometric forms.
(b)

Answer to Problem 46E
The product using the trigonometric form is z=0.732+2.73i
Explanation of Solution
Given info:
(√3+i)(1+i)
Formula used:
Let z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2) we have
z1z2=r1r2(cos(θ1+θ2)+isin(θ1+θ2))z1z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))
Calculation:
From part a we have (√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]
We have
z=(√3+i)(1+i)=[2(cosπ6+isinπ6)][√2(cosπ4+isinπ4)]z1=r1(cosθ1+isinθ1)=2(cosπ6+isinπ6)z2=r2(cosθ2+isinθ2)=√2(cosπ4+isinπ4)z=z1z2=r1r2(cos(θ1+θ2)+isin(θ1+θ2))z=2√2(cos(π6+π4)+isin(π6+π4))z=2√2(cos5π12+isin5π12)z=0.732+2.73i
The product using the trigonometric formis z=0.732+2.73i
Conclusion:
Thus,the product using the trigonometric form is z=0.732+2.73i
(c)
The product using the standard forms, and checkyour result with that of part (b).
(c)

Answer to Problem 46E
The product using standard form and trigonometric form are same.
Explanation of Solution
Given info:
(√3+i)(1+i)
Formula used:
We have the results
i2=−1
Calculation:
Product from part b is z=0.732+2.73i
The product using the standard forms is given by
z=(√3+i)(1+i)z=√3+√3i+i−i2=0.732+2.73i
Product using standard form and trigonometric form are same.
Conclusion:
Thus,the product using standard form and trigonometric form are same.
Chapter 6 Solutions
EBK PRECALCULUS W/LIMITS
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