Agnes Scott College
Larry Riddle, Agnes Scott College

Calculation using Complex numbers and Vectors

complex

Let \(\alpha = {e^{{{A}^ \circ}i}}\) (a complex number representing a counterclockwise rotation by A°) and \(\beta = {e^{{72^ \circ}i}}\) (a complex number representing counterclockwise rotation by 72°). The first segment can then be represented by the vector \(r\alpha \). Representing the other segments as vectors with the appropriate rotation, we can add the vectors to get a sum that must equal 1, that is

\[\begin{array}{l} r\alpha + r\alpha \beta + r\alpha + r\alpha {\beta ^{ - 2}} + r\alpha {\beta ^{ - 1}} + r\alpha = 1 \\ \\ \Rightarrow r\alpha \left( {3 + \beta + {\beta ^{ - 2}} + {\beta ^{ - 1}}} \right) = 1 \\ \\ \Rightarrow r{e^{Ai}} = \frac{1}{{3 + \beta + {\beta ^{ - 2}} + {\beta ^{ - 1}}}} \\ \end{array}\]

Evaluating this complex fraction numerically gives

\[r{e^{Ai}} = \frac{1}{{3 + \beta + {\beta ^{ - 2}} + {\beta ^{ - 1}}}} = 0.34844966{e^{{{11.81858573}^ \circ }i}}\]

This shows that r = 0.34845 and A = 11.81858573°.

To compute the exact value of r, we must be more careful with the calculation. Using DeMoivre's Theorem, we first get that

Trig Value
Details

\[\begin{align} 3 + \beta + {\beta ^{ - 2}} + {\beta ^{ - 1}} &= 3 + \left( {\cos {{72}^ \circ } + i\sin {{72}^ \circ }} \right) + \left( {\cos ( - {{144}^ \circ }) + i\sin ( - {{144}^ \circ })} \right) + \left( {\cos ( - {{72}^ \circ }) + i\sin ( - {{72}^ \circ })} \right) \\ &= \left( {3 + 2\cos {{72}^ \circ } + \cos {{144}^ \circ }} \right) - i\sin {144^ \circ } \\ &= \left( {3 + 2\cos {{72}^ \circ } - \cos {{36}^ \circ }} \right) - i\sin {36^ \circ } \\ &= \left( {3 + \frac{{\sqrt 5 - 1}}{2} - \frac{{1 + \sqrt 5 }}{4}} \right) - i \cdot \sqrt {\frac{{5 - \sqrt 5 }}{8}} \\ &= \frac{{9 + \sqrt 5 }}{4} - i \cdot \frac{{\sqrt {10 - 2\sqrt 5 } }}{4} \\ \end{align}\]

Now taking the reciprocal yields

\[r{e^{{A^ \circ }i}} = \frac{1}{{\frac{{9 + \sqrt 5 }}{4} - i \cdot \frac{{\sqrt {10 - 2\sqrt 5 } }}{4}}} = \frac{{9 + \sqrt 5 }}{{4\left( {6 + \sqrt 5 } \right)}} + i \cdot \frac{{\sqrt {10 - 2\sqrt 5 } }}{{4\left( {6 + \sqrt 5 } \right)}}\]

The absolute value of this complex number shows us that

\[r = \frac{1}{{\sqrt {6 + \sqrt 5 } }} = \sqrt {\frac{{6 - \sqrt 5 }}{{31}}} \]

There is really no advantage in trying to find an exact value for the angle A. But.....if you really need to know, see the derivation using geometry and trigonometry.