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Chapter 7: Problem 3

Write the following in rectangular form, \(z=a+i b\). a. \(4 e^{i \pi / 6}\) b. \(\sqrt{2} e^{5 i \pi / 4}\) c. \((1-i)^{100} .\)

Short Answer

Expert verified
a. \(2\sqrt{3} + 2i\), b. \(-1 + i\), c. \(-1\)

Step by step solution

01

Problem a: Convert \(4 e^{i \pi / 6}\) to Rectangular Form

From Euler's formula, this can be written as \(4(\cos(\pi / 6) + i \sin(\pi / 6))\). Then, using the values cos(\(\pi / 6\) = \(\sqrt{3}/ 2\), and \sin(\pi / 6) = 0.5, simplification gives \(2\sqrt{3} + 2i\).
02

Problem b: Convert \(\sqrt{2} e^{5 i \pi / 4}\) to Rectangular Form

Using Euler's formula, this can be written as \(\sqrt{2}(\cos(5\pi / 4) + i \sin(5 \pi / 4))\). Knowing that \cos(\(5 \pi / 4\)) = -\(\frac1{\sqrt2}\), and \sin(\(5 \pi / 4\)) = \(\frac1{\sqrt2}\), simplification gives \(-1 + i\).
03

Problem c: Convert \((1-i)^{100}\) to Rectangular Form

The first step is to write \(1 - i\) in polar form. This gives \(e^{7i\pi / 4}\), due to \(\cos(7 \pi / 4) = \sqrt{2}/ 2\) and \(\sin(7 \pi / 4) = - \sqrt{2}/ 2\). Raising this to the power 100 gives \(e^{175i\pi} = (e^{i\pi})^{175}\).\nThe 175 is an odd number, and we know that \(e^{i\pi} = -1\), so it can be simplified as \(-1\).

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Most popular questions from this chapter

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