Question #116825
Express −1 + i in polar form. Hence show that (−1 + i)16 is real and that 1/(−1 + i)6
is purely imaginary, giving the value for each.
1
Expert's answer
2020-05-25T20:31:07-0400

z=1+i=z(cosθ+isinθ)z=-1+i=|z|(cos\theta+isin\theta)

z=1+1=2|z|=\sqrt{1+1}=\sqrt{2}

cosθ=1/2,sinθ=1/2,θ=3π/4cos\theta=-1/\sqrt{2}, sin\theta=1/\sqrt{2}, \theta=3\pi/4

1+i=2(cos3π4+isin3π4)-1+i=\sqrt{2}(cos\frac {3\pi}{4}+isin\frac {3\pi}{4})


zn=zn(cos(nθ)+isin(nθ))z^n=|z|^n(cos(n\theta)+isin(n\theta))

(1+i)16=(2)16(cos(163π4)+isin(163π4)=(-1+i)^{16}=(\sqrt{2})^{16}(cos(\frac {16\cdot3\pi}{4})+isin(\frac {16\cdot3\pi}{4})=

=28(cos12π+isin12π)=256(1+0)=256=2^8(cos12\pi+isin12\pi)=256(1+0)=256


1(1+i)6=1(2)6(cos(63π4)+isin(63π4)=\frac {1}{(-1+i)^6}=\frac {1}{(\sqrt{2})^6(cos(\frac {6\cdot3\pi}{4})+isin(\frac {6\cdot3\pi}{4})}=

=123(cos(9π/2)+isin(9π/2))=18(0+i)=18i=i/8=\frac {1}{2^3(cos(9\pi/2)+isin(9\pi/2))}=\frac {1}{8(0+i)}=\frac {1}{8i}=-i/8


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