Complex Number Question (1 Viewer)

MrBeefJerky · Dec 26, 2013

Any hints/tips on how to solve this question?
If (x+iy)^n = a+ib, find a^2+b^2

Ikki · Dec 26, 2013

Polar form sounds nice, because you've got an 'n', demoivres can help you out.

cineti970128 · Dec 26, 2013

(x^2 + y^2)^n

Because a^2+b^2 is LHS's modulus squared.

asianese · Dec 26, 2013

integral95 · Dec 26, 2013

Another way is considering the conjugate $\\ (x-iy)^n = a-ib$

So there fore $(a+ib)(a-ib) = (x+iy)^n(x-iy)^n = \\ (a^2+b^2) = (x^2+y^2)^n$

Carrotsticks · Dec 26, 2013

MrBeefJerky said:
Any hints/tips on how to solve this question?
If (x+iy)^n = a+ib, find a^2+b^2

$a^2+b^2 = |a+ib|^2 = \left | (x+iy)^n \right |^2 = |x+iy|^{2n} = \left(|x+iy|^2\right)^n = \left ( x^2 + y^2 \right )^n$

asianese · Dec 26, 2013

Carrotsticks said:
$\\ a^2 +b^2 = |a+ib| = \left | (x+iy)^n \right | = |x+iy|^n = \left ( x^2 + y^2 \right )^n$

r u sure

$a^2+b^2 = |a+ib|^2 = \left | (x+iy)^n \right |^2 = |x+iy|^{2n} = \left(|x+iy|^2\right)^n = \left ( x^2 + y^2 \right )^n$

Carrotsticks · Dec 26, 2013

asianese said:
r u sure

$a^2+b^2 = |a+ib|^2 = \left | (x+iy)^n \right |^2 = |x+iy|^{2n} = \left(|x+iy|^2\right)^n = \left ( x^2 + y^2 \right )^n$

Thanks for that! I often forget that it is modulus squared, not the modulus =)

asianese · Dec 26, 2013

Carrotsticks said:
Thanks for that! I often forget that it is modulus squared, not the modulus =)

Pythagoras would not be impressed.

MrBeefJerky · Dec 26, 2013

Thank you guys!