Chapter 9: Q. 51 (page 592)

In Problems 41–52, write each expression in the standard form $a + b i$ .
$51 . {(\sqrt{2} - i)}^{6}$

Short Answer

Expert verified

${(\sqrt{2} - i)}^{6} = - 23 + 14.14 i$

Step by step solution

Step 1. Rewrite a+bi as r(cos∅+isin∅).

The magnitude of $(\sqrt{2} - i)$ is

role="math" localid="1646806173217" $\sqrt{{(\sqrt{2})}^{2} + {(- 1)}^{2}} = \sqrt{2 + 1} = \sqrt{3}$

This gives

$(\sqrt{2} - i) = \sqrt{3} (\frac{\sqrt{2}}{\sqrt{3}} - i \frac{1}{\sqrt{3}}) = \sqrt{3} (\cos 35.264 ° + i \sin 35.264 °)$

Step 2. Using De Moivre's Theorem.

If a complex number is $z = r (\cos \emptyset + i \sin \emptyset)$ then $z^{n} = r^{n} [\cos (n \emptyset) + i \sin (n \emptyset)]$ .

This gives

role="math" localid="1646806473235" ${(\sqrt{2} - i)}^{6} = {[\sqrt{3} (\cos 35.264 °) - i \sin (35.264 °)]}^{6} = {(\sqrt{3})^{6} [\cos (6 \times 35.264) - i \sin (6 \times 35.264)]} = 27 (\cos 211.584 ° - i \sin 211.584 °) = 27 (- 0.8518 - i (- 0.8518)) = - 23 + 14.14 i$

So, ${(\sqrt{2} - i)}^{6} = - 23 + 14.14 i$

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In Problems 41–52, write each expression in the standard form $a + b i$ .
$51 . {(\sqrt{2} - i)}^{6}$

Short Answer

Step by step solution

Step 1. Rewrite a+bi as r(cos∅+isin∅).

Step 2. Using De Moivre's Theorem.

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In Problems 41–52, write each expression in the standard form a+bi.51.(2-i)6

Short Answer

Step by step solution

Step 1. Rewrite a+bi as r(cos∅+isin∅).

Step 2. Using De Moivre's Theorem.

One App. One Place for Learning.

Most popular questions from this chapter

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In Problems 41–52, write each expression in the standard form $a + b i$ .
$51 . {(\sqrt{2} - i)}^{6}$