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    not related to OP but I also have a complex number question:
    (i) Expand and simplify (a + bi)^3 where a and b are real numbers, and i is  \sqrt{-1}

    I got b^3 i^3 + 3 a b^2 i^2 + 3 a^2 b i + a^3
    which simplified to  a^3 - 3ab^2 + (3 a^2 b - b^3)i

    (ii)Deduce that if (a +bi)^3 is real then either b=0 or b^2 = 3a^2

    I understand b=0, because that would make all of the terms with b as a factor, which includes all of the terms with i, 0. But I have no idea how to do the second one.
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    (Original post by Shipreck)
    not related to OP but I also have a complex number question:
    (i) Expand and simplify (a + bi)^3 where a and b are real numbers, and i is  \sqrt{-1}

    I got b^3 i^3 + 3 a b^2 i^2 + 3 a^2 b i + a^3
    which simplified to  a^3 - 3ab^2 + (3 a^2 b - b^3)i

    (ii)Deduce that if (a +bi)^3 is real then either b=0 or b^2 = 3a^2

    I understand b=0, because that would make all of the terms with b as a factor, which includes all of the terms with i, 0. But I have no idea how to do the second one.
    If it is real, then the imaginary part of the complex number  (a+bi)^{3} would be zero, i.e  b(3a^{2}-b^{2}) = 0 . This means that either  b = 0 or  3a^{2} - b^{2} = 0 giving the second required result.
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    (Original post by Shipreck)
    not related to OP but I also have a complex number question:
    (i) Expand and simplify (a + bi)^3 where a and b are real numbers, and i is  \sqrt{-1}

    I got b^3 i^3 + 3 a b^2 i^2 + 3 a^2 b i + a^3
    which simplified to  a^3 - 3ab^2 + (3 a^2 b - b^3)i

    (ii)Deduce that if (a +bi)^3 is real then either b=0 or b^2 = 3a^2

    I understand b=0, because that would make all of the terms with b as a factor, which includes all of the terms with i, 0. But I have no idea how to do the second one.
    You expanded correctly. Since you are convinced that b=0 will make it real, look at the imaginary part of the expansion and make it equal 0. Divide that equation through by b (assuming its not 0) and there you have it. The post above me would get you both solutions at once by means of factorising out the b rather than dividing which is more useful than by looking at the expanded and factorised forms separately.
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    (Original post by Shipreck)
    not related to OP but I also have a complex number question:
    (i) Expand and simplify (a + bi)^3 where a and b are real numbers, and i is  \sqrt{-1}

    I got b^3 i^3 + 3 a b^2 i^2 + 3 a^2 b i + a^3
    which simplified to  a^3 - 3ab^2 + (3 a^2 b - b^3)i

    (ii)Deduce that if (a +bi)^3 is real then either b=0 or b^2 = 3a^2

    I understand b=0, because that would make all of the terms with b as a factor, which includes all of the terms with i, 0. But I have no idea how to do the second one.
    If (a +bi)^3 is real, then the imaginary part is 0 i.e. in a^3 - 3ab^2 + (3 a^2 b - b^3)i, (3a^2b-b^3) = 0.

    Hence we can factorise the expression: b(3a^2-b^2) = 0. I'm sure you can now see we get b=0 and b^2=3a^2.
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    (Original post by RDKGames)
    You expanded correctly. Since you are convinced that b=0 will make it real, look at the imaginary part of the expansion and make it equal 0. Divide that equation through by b (assuming its not 0) and there you have it. The post above me would get you both solutions at once by means of factorising out the b rather than dividing which is more useful than by looking at the expanded and factorised forms separately.
    (Original post by NeverLucky)
    If (a +bi)^3 is real, then the imaginary part is 0 i.e. in a^3 - 3ab^2 + (3 a^2 b - b^3)i, (3a^2b-b^3) = 0.

    Hence we can factorise the expression: b(3a^2-b^2) = 0. I'm sure you can now see we get b=0 and b^2=3a^2.

    (Original post by AMarques)
    If it is real, then the imaginary part of the complex number  (a+bi)^{3} would be zero, i.e  b(3a^{2}-b^{2}) = 0 . This means that either  b = 0 or  3a^{2} - b^{2} = 0 giving the second required result.
    Thank you. I don't know how I didn't see that. I'm such an idiot XD.
 
 
 
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