Complex Addition/Examples/Travel 1

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Example of Complex Addition

A man travels:

$12$ kilometres northeast

then:

$20$ kilometres $30 \degrees$ west of north

then:

$18$ kilometres $60 \degrees$ south of west.


Assuming the curvature of the Earth to be negligible at this scale, at the end of this travel, he is $14.7$ kilometres in a direction $45 \degrees 49'$ west of north from his starting point.


Proof 1

Let the route of the traveller be embedded in the complex plane.


Travel Example.png


Let $P$ be the final location of the traveller.


The given directions can be translated into absolute arguments thus:

Northeast is $45 \degrees$ north of east, and therefore an argument of $45 \degrees$
$30 \degrees$ west of north is $30 + 90 \degrees$ north of east, and therefore an argument of $120 \degrees$
$60 \degrees$ south of west is $60 + 180 \degrees$ north of east, and therefore an argument of $240 \degrees$.


Thus the problem reduces to finding the sum:

\(\ds P\) \(=\) \(\ds 12 \cis 45 \degrees + 20 \cis 120 \degrees + 18 \cis 240 \degrees\)
\(\ds \) \(=\) \(\ds \paren {12 \cos 45 \degrees + 20 \cos 120 \degrees + 18 \cos 240 \degrees} + i \paren {12 \sin 45 \degrees + 20 \sin 120 \degrees + 18 \sin 240 \degrees}\)
\(\ds \) \(=\) \(\ds \paren {12 \frac {\sqrt 2} 2 + 20 \paren {-\dfrac 1 2} + 18 \paren {-\dfrac 1 2} } + i \paren {12 \sin 45 \degrees + 20 \sin 120 \degrees + 18 \sin 240 \degrees}\) Cosine of $45 \degrees$, Cosine of $120 \degrees$, Cosine of $240 \degrees$
\(\ds \) \(=\) \(\ds \paren {12 \frac {\sqrt 2} 2 + 20 \paren {-\dfrac 1 2} + 18 \paren {-\dfrac 1 2} } + i \paren {12 \frac {\sqrt 2} 2 + 20 \frac {\sqrt 3} 2 + 18 \paren {-\frac {\sqrt 3} 2} }\) Sine of $45 \degrees$, Sine of $120 \degrees$, Sine of $240 \degrees$
\(\ds \) \(=\) \(\ds \paren {6 \sqrt 2 - 19} + i \paren {6 \sqrt 2 + \sqrt 3}\) after algebra

Then:

\(\ds R \cis \theta\) \(=\) \(\ds \paren {6 \sqrt 2 - 19} + i \paren {6 \sqrt 2 + \sqrt 3}\)
\(\ds \leadsto \ \ \) \(\ds R\) \(=\) \(\ds \sqrt {\paren {6 \sqrt 2 - 19}^2 + \paren {6 \sqrt 2 + \sqrt 3}^2}\)
\(\ds \) \(\approx\) \(\ds 14.7\)
\(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \cos^{-1} \dfrac {6 \sqrt 2 - 19} {14.7}\)
\(\ds \) \(\approx\) \(\ds \map {\cos^{-1} } {- 0 \cdotp 717}\)
\(\ds \) \(\approx\) \(\ds 135 \degrees 49'\)
\(\ds \) \(\approx\) \(\ds 45 \degrees 49' \text { west of north}\)

$\blacksquare$


Proof 2

By plotting the points in a graphics package, or on paper with a ruler and protractor:


Travel Example.png


$\blacksquare$