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Express complex number $(i^{10} + i^{-7})^{11}$ in Euler form, where $r > 0$ and $ - \pi < \theta \le \pi$, where $\theta$ is in the form of $k\pi$

I convert $(i^{10} + i^{-7})^{11}$ to $(-1 + i)^{11}$

In Euler form, $ (\sqrt{2} e^{(3 \pi/4) i})^{11}$

$e^{mi + 2k \pi i} = e^{mi}$

$e^{8mi + 1/4 \pi i} = e^{1/4 \pi i}$

Why is the answer $32 \sqrt{2} e^{(\pi / 4)i}$

How do I get $32\sqrt{2}$ ?

Kevin Dietrich
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user307640
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    There is the $(\sqrt{2})^{11}$ and $(\sqrt{2})^{11}=32\sqrt{2}$:$$\begin{align}(\sqrt{2} e^{(3/4\pi)i})^{11}&=(\sqrt{2})^{11}(e^{(3/4\pi)i})^{11}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=(2^{1/2})^{11} \cdot (e^{(3/4\pi) i})^{11}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=2^{11/2}\cdot e^{11(3/4\pi) i}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=2^{5.5}\cdot e^{11\cdot(3/4\pi)i}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=2^{5+1/2}\cdot e^{11(3/4\pi)i}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=2^{5}\cdot 2^{1/2}\cdot e^{11(3/4\pi)i}\(\sqrt{2}\cdot e^{(3/4\pi)i})^{11}&=32\cdot\sqrt{2}\cdot e^{11(3/4\pi)i}\\end{align}$$ – Kevin Dietrich Oct 24 '22 at 15:28

1 Answers1

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There is the $(\sqrt{2})^{11}$ in $\left( \sqrt{2} \cdot e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11} = \left( \sqrt{2} \right)^{11} \cdot \left( e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11}$ and $(\sqrt{2})^{11} = 32 \cdot \sqrt{2}$...

I can also show it to you step by step: $$ \begin{align*} x &= \left( \mathrm{i}^{10} + \mathrm{i}^{-7} \right)^{11}\\ x &= \left( -1 + \mathrm{i} \right)^{11}\\ x &= \left( \sqrt{(-1)^{2} + (1)^{2}} \cdot e^{\arctan2(1, -1) \cdot \mathrm{i}} \right)^{11}\\ x &= \left( \sqrt{1 + 1} \cdot e^{\arctan2(1, -1) \cdot \mathrm{i}} \right)^{11}\\ x &= \left( \sqrt{2} \cdot e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11}\\ x &= \left( \sqrt{2} \right)^{11} \cdot \left( e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11}\\ x &= \left( 2^{\frac{1}{2}} \right)^{11} \cdot \left( e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11}\\ x &= \left( 2^{0.5} \right)^{11} \cdot \left( e^{\frac{3}{4} \cdot \pi \cdot \mathrm{i}} \right)^{11}\\ x &= 2^{11 \cdot 0.5} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 2^{5.5} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 2^{5 + 0.5} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 2^{5} \cdot 2^{0.5} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 2^{5} \cdot 2^{\frac{1}{2}} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 32 \cdot \sqrt{2} \cdot e^{11 \cdot \frac{3}{4} \cdot \pi \cdot \mathrm{i}}\\ x &= 32 \cdot \sqrt{2} \cdot e^{\left( \frac{1}{4} \cdot \pi + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}}\\ x &= 32 \cdot \sqrt{2} \cdot e^{\frac{1}{4} \cdot \pi \cdot \mathrm{i} + 2 \cdot k \cdot \pi \cdot \mathrm{i}}\\ x &= 32 \cdot \sqrt{2} \cdot e^{\frac{1}{4} \cdot \pi \cdot \mathrm{i}}\\ \end{align*} $$

Kevin Dietrich
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