The complex number zz has modulus 2 and argument π8\frac{\pi}{8}. It is also given that w=1+iw = 1 + \mathrm{i}.

  1. (i)

    Given that nn is an integer, find znw∗\frac{z^n}{w^*} in terms of nn, giving your answer in the form reiθr\mathrm{e}^{\mathrm{i}\theta}.

  2. (ii)

    Hence, find the smallest two positive integers nn such that znw∗\frac{z^n}{w^*} is real and negative.

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Worked solution

2(i)

1. Express $z^n$ in exponential form.

z=2eiπ8z = 2\mathrm{e}^{\mathrm{i}\frac{\pi}{8}}
zn=2neinπ8z^n = 2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}

2. Express the conjugate $w^*$ in exponential form.

w=1+iw = 1 + \mathrm{i}
∣w∣=12+12=2|w| = \sqrt{1^2 + 1^2} = \sqrt{2}
arg⁡(w)=tan⁡−1(1)=π4\arg(w) = \tan^{-1}(1) = \frac{\pi}{4}
w=2eiπ4w = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}
w∗=2e−iπ4w^* = \sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}

3. Evaluate the quotient $\frac{z^n}{w^*}$.

znw∗=2neinπ82e−iπ4\frac{z^n}{w^*} = \frac{2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}}{\sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}
=2n212ei(nπ8−(−π4))= \frac{2^n}{2^{\frac{1}{2}}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} - \left(-\frac{\pi}{4}\right)\right)}
=2n−12ei(nπ8+2π8)= 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} + \frac{2\pi}{8}\right)}
=2n−12ei(n+2)π8= 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\frac{(n+2)\pi}{8}}

Answer

2n−12ei(n+2)π82^{n-\frac{1}{2}}\mathrm{e}^{\mathrm{i}\frac{(n+2)\pi}{8}}

2(ii)

1. Determine the polar form of the expression using the result from part (i).

From part (i), we establish the exponential forms:
z=2eiπ8  ⟹  zn=2neinπ8z = 2\mathrm{e}^{\mathrm{i}\frac{\pi}{8}} \implies z^n = 2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}
w=1+i=2eiπ4  ⟹  w∗=2e−iπ4w = 1 + \mathrm{i} = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} \implies w^* = \sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}
znw∗=2n2ei(nπ8−(−π4))=2n−12ei(nπ8+π4)\frac{z^n}{w^*} = \frac{2^n}{\sqrt{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} - \left(-\frac{\pi}{4}\right)\right)} = 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} + \frac{\pi}{4}\right)}

2. Set up the condition for the complex number to be real and negative.

For a complex number to be real and negative, its argument must be an odd multiple of π\pi. Therefore, we equate the argument to (2k+1)π(2k + 1)\pi where k∈Zk \in \mathbb{Z}:
nπ8+π4=(2k+1)π\frac{n\pi}{8} + \frac{\pi}{4} = (2k + 1)\pi

3. Solve the argument equation for the smallest two positive integers $n$.

Dividing the entire equation by π\pi and multiplying by 88:
n+2=8(2k+1)n + 2 = 8(2k + 1)
n=16k+8−2n = 16k + 8 - 2
n=16k+6n = 16k + 6
Since we require nn to be a positive integer (n>0n > 0), we take the smallest non-negative integer values for kk:
When k=0k = 0, n=16(0)+6=6n = 16(0) + 6 = 6
When k=1k = 1, n=16(1)+6=22n = 16(1) + 6 = 22

Answer

n=6n = 6 and n=22n = 22

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