let z=r(cosθ+isinθ) be any complex number & n be any positive integer. Then z^{n} =r^{n} (cos(nθ)+isin(nθ)) or equivelently, [r^{n} , nθ]. Prove it in special cases n=2,3,4
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Originally Posted by franios Prove it in special cases n=2 So, why don't you do this?
I'm just confused, do I have to just plug in n? What about theta?
Do you know how to prove that for n = 2 it is the case that for all x and y, (x + y)^n = x^n + 2xy + y^n? Now prove that for n = 2 it is the case that for all r and θ, (r(cosθ+isinθ))^n = r^{n}(cos(nθ)+isin(nθ)).
Originally Posted by franios I'm just confused, do I have to just plug in n? What about theta? $\displaystyle \begin{align*}z\cdot z &=r(cos(t)+i\sin(t))r(cos(t)+i\sin(t))\\&=r^2(\cos ^2(t)-\sin^2(t))+i(\cos(t)\sin(t))\\&=r^2(cos(2t)+i\sin( 2t))\end{align*}$ Now $\displaystyle z^3=z^2\cdot z$
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