Solution 3.3:1d

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Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write
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<center> [[Image:3_3_1d-1(2).gif]] </center>
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<math>1+i\sqrt{3}</math> and <math>\text{1}+i</math> in polar form right from the beginning and to carry out all calculations in polar form.
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<center> [[Image:3_3_1d-2(2).gif]] </center>
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[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]
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<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>
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We obtain
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{{Displayed math||<math>\begin{align}
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1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
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1+i &= \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)
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\end{align}</math>}}
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and
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{{Displayed math||<math>\begin{align}
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\frac{1+i\sqrt{3}}{1+i}
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&= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)}{\sqrt{2}\Bigl(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\Bigr)}\\[5pt]
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&= \frac{2}{\sqrt{2}}\Bigl(\cos\Bigl(\frac{\pi}{3}-\frac{\pi}{4}\Bigr) + i\sin\Bigl(\frac{\pi}{3}-\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]
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&= \sqrt{2}\Bigl(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\Bigr)\,\textrm{.}
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\end{align}</math>}}
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Finally, de Moivre's formula gives
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{{Displayed math||<math>\begin{align}
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\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}
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&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]
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&= 2^{(1/2)\cdot 12}(\cos\pi + i\sin\pi)\\[5pt]
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&= 2^6\cdot (-1+i\cdot 0)\\[5pt]
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&= -64\,\textrm{.}
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\end{align}</math>}}

Current revision

Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write \displaystyle 1+i\sqrt{3} and \displaystyle \text{1}+i in polar form right from the beginning and to carry out all calculations in polar form.

Image:3_3_1_d.gif Image:3_3_1_d_text.gif

We obtain

\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\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr) \end{align}

and

\displaystyle \begin{align}

\frac{1+i\sqrt{3}}{1+i} &= \frac{2\Bigl(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\Bigr)}{\sqrt{2}\Bigl(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\Bigr)}\\[5pt] &= \frac{2}{\sqrt{2}}\Bigl(\cos\Bigl(\frac{\pi}{3}-\frac{\pi}{4}\Bigr) + i\sin\Bigl(\frac{\pi}{3}-\frac{\pi}{4}\Bigr)\Bigr)\\[5pt] &= \sqrt{2}\Bigl(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\Bigr)\,\textrm{.} \end{align}

Finally, de Moivre's formula gives

\displaystyle \begin{align}

\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12} &= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt] &= 2^{(1/2)\cdot 12}(\cos\pi + i\sin\pi)\\[5pt] &= 2^6\cdot (-1+i\cdot 0)\\[5pt] &= -64\,\textrm{.} \end{align}

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