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Express in the form of $a+ib$. $$(1-i)^9 \bigg(1-\dfrac {1}{i^3}\bigg)^9$$


My Attempt:

$$\begin{align} &=(1-i)^9\bigg(1-\dfrac {1}{i^3}\bigg) \\ \\ &=(1-i)^9\bigg(1+\dfrac {1}{i}\bigg) \\ \\&=(1-i)^9\cdot(1-i)^9 \\ \\ &=(1-i)^{18} \end{align}$$

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Hints:

  • $\require{cancel}(1-i)^2=\bcancel{1}-2i+\bcancel{i^2}=-2i$

  • $1-1/i^3=1 - i / i^4=1 - i$

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  • $\begingroup$ what does this lead to? $\endgroup$
    – pi-π
    Jan 10 '18 at 7:18
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    $\begingroup$ @blue_eyed_... Leads to $\,(-2i)^{9}=-2^9 \cdot i^9 = \ldots\,$ $\endgroup$
    – dxiv
    Jan 10 '18 at 7:21
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Use polar form for $1-i$ which is $\sqrt2e^{-i \pi/4}$.

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    $\begingroup$ Don't know why someone downvoted this answer without at least leaving a comment. Fixing the obvious typo would have been more appropriate, by far. $\endgroup$
    – dxiv
    Jan 10 '18 at 7:36
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    $\begingroup$ Thanks for fixing it. $\endgroup$
    – Hamid Enki
    Jan 13 '18 at 3:37
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First of all calculate the first power of $(1-i)$: $$(1-i)^2=(-i)^2-2i+1=-1-2i+1=-2i\tag{1}$$ $$(1-i)^3=(-2i)\cdot(1-i)=-2i-2i^2=-2-2i=-2(1-i)\tag{2}$$

We also know that: $$\frac 1i=\frac 1i\cdot\frac ii=\frac i{-1}=-i\tag{3}$$

Now apply $(1)$,$(2)$ and $(3)$ to your problem and we get:

$$(1-i)^9\bigg(1-\dfrac {1}{i^3}\bigg)^9=\\=((1-i)^3)^3\bigg(1-\dfrac {1}{i^3}\bigg)^9\overbrace{=}^{(2)}(-2(1-i))^3\bigg(1+\dfrac {1}{i}\bigg)^9=\\=-8(1-i)^3\bigg(1+\dfrac {1}{i}\bigg)^9\overbrace{=}^{(2)}-8(-2(1-i))\bigg(1+\dfrac {1}{i}\bigg)^9=\\=16(1-i)\bigg(1+\dfrac {1}{i}\bigg)^9\overbrace{=}^{(3)}16(1-i)(1-i)^9=16(1-i)^{10}=\\=16((1-i)^2)^5\overbrace{=}^{(1)}16\cdot(-2i)^5=16\cdot(-32i)=\color{red}{-512i=(1-i)^{18}}$$

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So... finish it.

$(1-i)^9 (1-\frac 1{i^3})^9=$

$(1-i)^9(1 - \frac 1i\frac 1{i^2})^9 = (1-i)^9(1+\frac 1i)^9$

$= (1-i)^9(1+\frac 1i*\frac ii)^9 = (1-i)^9(1+\frac i{-1})^9 = (1-i)^9(1-i)^9$

$= (1-i)^{18}$.

So finish it.

$(1-i)^2 = 1^2 - 2i + i^2 = -2i$

$(1-i)^4 = (-2i)^2 = (-2)^2i^2 = -4$

$(1-i)^8 = (-4)^2 = 16$

$(1-i)^{16} = 16^2 = (2^4)^2 = 2^8$.

$(1-i)^{18} = (1-i)^{16}(1-i)^2 = 2^8*-2i = -2^9i = 0 - 512i$

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Using Hamid's hint:

Polar form: $re^{i\phi} :$

$1-i = √2e^{-i(π/4)} .$

$(1-i)^{18} = (√2)^{18} e^{-i(π/4)18}=$

$2^9e^{-i(π/4)2} = 2^9e^{-i(π/2)} = 2^9(-i).$

Answer: $0 - 2^9i.$

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