Recent content by angus.myers

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    Complex Numbers: nth Root Proof

    Haha awesome! Thank you very much for your help! Much appreciated!
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    Complex Numbers: nth Root Proof

    Wow! I'm an idiot! haha, Thank you for that! I didn't think of it that way and I think I was trying to over complicate things! so z = (a +ib)^n and z(bar) = (a - ib)^n when asked: Prove that if the complex number z is an nth root of the real number x then z is also an nth root of x. means...
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    Complex Numbers: nth Root Proof

    I think here it's more a case of what the complex conjugate of z actually is. I think that is where I am having difficulties! I've never touched on complex numbers before, I'm trying to read up on it but am obviously having basic ground work issues. How would one go about finding the complex...
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    Complex Numbers: nth Root Proof

    Did you ever manage to get this problem solved? z = a - ib and z= a+ib are complex conjugates are they not? Where both a and b are real numbers. so in this case, if z = n(sqrt)ia then the complex conjugate is = n(sqrt)-ia is this correct?
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