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प्रश्न
Solve the equation z3 + 27 = 0
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उत्तर
z3 + 27 = 0
⇒ z3 = – 27
⇒ z3 = `3^3(- 1)`
⇒ z = `3(-1)^(1/3)`
z = `3[cos(2"k"pi + pi) + "i" sin 2"k"pi + pi]^(1/3)`
= `3[cos((2"k"pi + pi)/3) + "i" sin((2"k"pi + pi)/3)]`
k = 0, 1, 2
k = 0, z = `3[cos pi/3 + "i" sin pi/3]`
k = 1, z = `3[cos (3pi)/3 + "i"sin (3pi)/3]` = – 3
k = 2, z = `3[cos (5pi)/3 + "i" sin (5pi)/3]`
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