Solution 3.3:1e
From Förberedande kurs i matematik 2
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| - | {{ | + | Some parts of the quotient have rather high exponents and this indicates that we ought to use polar form for the calculation. |
| - | <center> [[Image: | + | |
| - | {{ | + | First, we write <math>1+i\sqrt{3}</math>, <math>1-i</math> and <math>\sqrt{3}-i</math> in polar form. |
| - | {{ | + | |
| - | < | + | <center>[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]]</center> |
| - | {{ | + | |
| + | This shows that | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | 1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\,,\\[5pt] | ||
| + | 1-i &= \sqrt{2}\Bigl(\cos\Bigl(-\frac{\pi}{4}\Bigr) + i\sin\Bigl(-\frac{\pi}{4}\Bigr)\Bigr)\,,\\[5pt] | ||
| + | \sqrt{3}-i &= 2\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Now, with the help of de Moivre's formula, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{\bigl(1+i\sqrt{3}\,\bigr)(1-i)^8}{\bigl(\sqrt{3}-i\bigr)^9} | ||
| + | &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\Bigl(\sqrt{2}\Bigl(\cos\Bigl(-\dfrac{\pi}{4}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)^8}{\Bigl( 2\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)^9}\\[5pt] | ||
| + | &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\bigl(\sqrt{2}\,\bigr)^8\Bigl(\cos\Bigl(8\cdot\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr) + i\sin\Bigl(8\cdot\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)}{2^{9}\Bigl(\cos\Bigl(9\cdot\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr) + i\sin\Bigl(9\cdot\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)}\\[5pt] | ||
| + | &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\cdot 2^{(1/2)\cdot 8}\bigl(\cos (-2\pi) + i\sin (-2\pi)\bigr)}{2^9\Bigl(\cos\Bigl(-\dfrac{3\pi}{2} \Bigr) + i\sin\Bigl(-\dfrac{3\pi}{2}\Bigr)\Bigr)}\\[5pt] | ||
| + | &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\cdot 2^4( 1+i\cdot 0)}{2^9\Bigl(\cos\Bigl(-\dfrac{3\pi}{2}\Bigr) + i\sin\Bigl(-\dfrac{3\pi}{2}\Bigr)\Bigr)}\\[5pt] | ||
| + | &= \frac{2\cdot 2^4}{2^9}\Bigl(\cos\Bigl(\frac{\pi}{3}-\Bigl(-\frac{3\pi}{2}\Bigr) \Bigr) + i\sin\Bigl(\frac{\pi}{3}-\Bigl(-\frac{3\pi}{2}\Bigr)\Bigr)\Bigr)\\[5pt] | ||
| + | &= \frac{2^5}{2^9}\Bigl(\cos\Bigl(\frac{\pi}{3}+\frac{3\pi}{2}\Bigr) + i\sin\Bigl(\frac{\pi}{3}+\frac{3\pi}{2}\Bigr)\Bigr)\\[5pt] | ||
| + | &= \frac{1}{2^4}\Bigl(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\Bigr)\\[5pt] | ||
| + | &= \frac{1}{16}\Bigl(\cos\frac{12\pi-\pi}{6} + i\sin\frac{12\pi-\pi}{6}\Bigr)\\[5pt] | ||
| + | &= \frac{1}{16}\Bigl(\cos\Bigl(2\pi-\frac{\pi}{6}\Bigr) + i\sin\Bigl(2\pi-\frac{\pi}{6}\Bigr)\Bigr)\\[5pt] | ||
| + | &= \frac{1}{16}\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\\[5pt] | ||
| + | &= \frac{1}{16}\Bigl(\frac{\sqrt{3}}{2}-\frac{i}{2}\Bigr)\\[5pt] | ||
| + | &= \frac{1}{32}\bigl(\sqrt{3}-i\bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
Some parts of the quotient have rather high exponents and this indicates that we ought to use polar form for the calculation.
First, we write \displaystyle 1+i\sqrt{3}, \displaystyle 1-i and \displaystyle \sqrt{3}-i in polar form.
This shows that
| \displaystyle \begin{align}
1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\,,\\[5pt] 1-i &= \sqrt{2}\Bigl(\cos\Bigl(-\frac{\pi}{4}\Bigr) + i\sin\Bigl(-\frac{\pi}{4}\Bigr)\Bigr)\,,\\[5pt] \sqrt{3}-i &= 2\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\,\textrm{.} \end{align} |
Now, with the help of de Moivre's formula,
| \displaystyle \begin{align}
\frac{\bigl(1+i\sqrt{3}\,\bigr)(1-i)^8}{\bigl(\sqrt{3}-i\bigr)^9} &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\Bigl(\sqrt{2}\Bigl(\cos\Bigl(-\dfrac{\pi}{4}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)^8}{\Bigl( 2\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr) + i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)^9}\\[5pt] &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\bigl(\sqrt{2}\,\bigr)^8\Bigl(\cos\Bigl(8\cdot\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr) + i\sin\Bigl(8\cdot\Bigl(-\dfrac{\pi}{4}\Bigr)\Bigr)\Bigr)}{2^{9}\Bigl(\cos\Bigl(9\cdot\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr) + i\sin\Bigl(9\cdot\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)\Bigr)}\\[5pt] &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\cdot 2^{(1/2)\cdot 8}\bigl(\cos (-2\pi) + i\sin (-2\pi)\bigr)}{2^9\Bigl(\cos\Bigl(-\dfrac{3\pi}{2} \Bigr) + i\sin\Bigl(-\dfrac{3\pi}{2}\Bigr)\Bigr)}\\[5pt] &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)\cdot 2^4( 1+i\cdot 0)}{2^9\Bigl(\cos\Bigl(-\dfrac{3\pi}{2}\Bigr) + i\sin\Bigl(-\dfrac{3\pi}{2}\Bigr)\Bigr)}\\[5pt] &= \frac{2\cdot 2^4}{2^9}\Bigl(\cos\Bigl(\frac{\pi}{3}-\Bigl(-\frac{3\pi}{2}\Bigr) \Bigr) + i\sin\Bigl(\frac{\pi}{3}-\Bigl(-\frac{3\pi}{2}\Bigr)\Bigr)\Bigr)\\[5pt] &= \frac{2^5}{2^9}\Bigl(\cos\Bigl(\frac{\pi}{3}+\frac{3\pi}{2}\Bigr) + i\sin\Bigl(\frac{\pi}{3}+\frac{3\pi}{2}\Bigr)\Bigr)\\[5pt] &= \frac{1}{2^4}\Bigl(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\Bigr)\\[5pt] &= \frac{1}{16}\Bigl(\cos\frac{12\pi-\pi}{6} + i\sin\frac{12\pi-\pi}{6}\Bigr)\\[5pt] &= \frac{1}{16}\Bigl(\cos\Bigl(2\pi-\frac{\pi}{6}\Bigr) + i\sin\Bigl(2\pi-\frac{\pi}{6}\Bigr)\Bigr)\\[5pt] &= \frac{1}{16}\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\\[5pt] &= \frac{1}{16}\Bigl(\frac{\sqrt{3}}{2}-\frac{i}{2}\Bigr)\\[5pt] &= \frac{1}{32}\bigl(\sqrt{3}-i\bigr)\,\textrm{.} \end{align} |
