Solution 3.3:4a
From Förberedande kurs i matematik 2
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| - | {{ | + | This is a typical binomial equation which we solve in polar form. |
| - | < | + | |
| - | {{ | + | We write |
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] | ||
| + | i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | and, on using de Moivre's formula, the equation becomes | ||
| + | |||
| + | {{Displayed math||<math>r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.}</math>}} | ||
| + | |||
| + | Both sides are equal when | ||
| + | |||
| + | {{Displayed math||<math>\left\{\begin{align} | ||
| + | r^2 &= 1\,,\\[5pt] | ||
| + | 2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),} | ||
| + | \end{align}\right.</math>}} | ||
| + | |||
| + | which gives that | ||
| + | |||
| + | {{Displayed math||<math>\left\{\begin{align} | ||
| + | r &= 1\,,\\[5pt] | ||
| + | \alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).} | ||
| + | \end{align}\right.</math>}} | ||
| + | |||
| + | When <math>n=0</math> and <math>n=1</math>, we get two different arguments for | ||
| + | <math>\alpha</math>, whilst different values of <math>n</math> only give these arguments plus/minus a multiple of <math>2\pi</math>. | ||
| + | |||
| + | The solutions to the equation are | ||
| + | |||
| + | {{Displayed math||<math>z=\left\{\begin{align} | ||
| + | &1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt] | ||
| + | &1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr) | ||
| + | \end{align}\right. | ||
| + | = | ||
| + | \left\{\begin{align} | ||
| + | &\frac{1+i}{\sqrt{2}}\,,\\[5pt] | ||
| + | &-\frac{1+i}{\sqrt{2}}\,\textrm{.} | ||
| + | \end{align}\right.</math>}} | ||
Current revision
This is a typical binomial equation which we solve in polar form.
We write
| \displaystyle \begin{align}
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,, \end{align} |
and, on using de Moivre's formula, the equation becomes
| \displaystyle r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.} |
Both sides are equal when
| \displaystyle \left\{\begin{align}
r^2 &= 1\,,\\[5pt] 2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),} \end{align}\right. |
which gives that
| \displaystyle \left\{\begin{align}
r &= 1\,,\\[5pt] \alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).} \end{align}\right. |
When \displaystyle n=0 and \displaystyle n=1, we get two different arguments for \displaystyle \alpha, whilst different values of \displaystyle n only give these arguments plus/minus a multiple of \displaystyle 2\pi.
The solutions to the equation are
| \displaystyle z=\left\{\begin{align}
&1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt] &1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr) \end{align}\right. = \left\{\begin{align} &\frac{1+i}{\sqrt{2}}\,,\\[5pt] &-\frac{1+i}{\sqrt{2}}\,\textrm{.} \end{align}\right. |
