View source for Mathematical Methods in the Physical Sciences/Chapter 2/Section 4

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{{:Mathematical Methods in the Physical Sciences/Chapter 2/section sidebar}}
== Notes ==

Complex conjugate: <math>\overline{z} = x-yi = re^{-i\theta}</math>
* This is the mirrored reflection about the real axis
* Either flip the sign of the imaginary part or just take the negative angle.

<math>|z| = \mod z = r = \sqrt{x^2 + y^2}</math>

When you use <math>\theta = \arctan (y/x)</math>, keep in mind that for pure imaginary numbers, <math>\theta = \pm {\pi \over 2}</math>. It's better to find the angle with geometry than arctan.

== Problems ==

I'm not going to graph these. It's all rather simple anyway.

<math>
\begin{array}{c|c|c|c|c}
z & x & y & r & \theta \\
\hline

1 + i & 1 & 1 & \sqrt{2} & {\pi \over 4} \\
&&&&\\
\overline{1 + i} & 1 & -1 & \sqrt{2} & -{\pi \over 4} \\
&&&&\\
i - 1 & -1 & 1 & \sqrt{2} & {3 \pi \over 4} \\
&&&&\\
\overline{i - 1} & -1 & -1 & \sqrt{2} & -{3 \pi \over 4} \\
&&&&\\
1 - i\sqrt{3} & 1 & -\sqrt{3} & 2 & -{\pi \over 3} \\
&&&&\\
\overline{1 - i\sqrt{3}} & 1 & \sqrt{3} & 2 & {\pi \over 3} \\
&&&&\\
-\sqrt{3} + i & -\sqrt{3} & 1 & 2 & {5 \pi \over 6} \\
&&&&\\
\overline{-\sqrt{3} + i} & -\sqrt{3} & -1 & 2 & -{5 \pi \over 6} \\
&&&&\\
2i & 0 & 2 & 2 & {\pi \over 2} \\
&&&&\\
\overline{2i} & 0 & -2 & -{\pi \over 2} \\
&&&&\\
-4i & 0 & -4 & 4 & -{\pi \over 2} \\
&&&&\\
\overline{-4i} & 0 & 4 & 4 & {\pi \over 2} \\
&&&&\\
-1 & -1 & 0 & 1 & \pi \\
&&&&\\
\overline{-1} & -1 & 0 & 1 & (-)\pi \\
&&&&\\
3 & 3 & 0 & 3 & 0 \\
&&&&\\
\overline{3} & 3 & 0 & 3 & (-)0 \\
&&&&\\
2i - 2 & -2 & 2 & \sqrt{8} & {3 \pi \over 4} \\
&&&&\\
\overline{2i - 2} & -2 & -22 & \sqrt{8} & -{3 \pi \over 4} \\
&&&&\\
2-2i & 2 & -2 & \sqrt{8} & -{\pi \over 4} \\
&&&&\\
\overline{2-2i} & 2 & 2 & \sqrt{8} & {\pi \over 4} \\
&&&&\\
2\left(\cos {\pi \over 6} + i \sin {\pi \over 6}\right) & \sqrt{3} & 1 & 2 & {\pi \over 6} \\
&&&&\\
\overline{2\left(\cos {\pi \over 6} + i \sin {\pi \over 6}\right)} & \sqrt{3} & -1 & 2 & -{\pi \over 6} \\
&&&&\\
4\left(\cos {2\pi \over 3} - i \sin {2\pi \over 3}\right) & -2 & -2\sqrt{3} & 4 & -{2\pi \over 3} \\
&&&&\\
\overline{4\left(\cos {2\pi \over 3} - i \sin {2\pi \over 3}\right)} & -2 & 2\sqrt{3} & 4 & {2\pi \over 3} \\
&&&&\\
\cos {3\pi \over 2} + i \sin {3\pi \over 2} & 0 & -1 & 1 & -{\pi \over 2} \\
&&&&\\
\overline{\cos {3\pi \over 2} + i \sin {3\pi \over 2}} & 0 & 1 & 1 & {\pi \over 2} \\
&&&&\\



\end{array}
</math>