Simplifying an expression involving complex numbers.

Dec 2009
180
19
The exercise is:
write the expression in polar form: \(\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}\)
Here is my work:
\(\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}\)
that is


\(\displaystyle \frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})(2\sqrt{3}-2i)}\)

and that becomes:


\(\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}\)


and then


\(\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}\)

\(\displaystyle \frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}\)

anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
The exercise is:
write the expression in polar form: \(\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}\)
Here is my work:
\(\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}\)
that is


\(\displaystyle \frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})(2\sqrt{3}-2i)}\)

and that becomes:


\(\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}\)


and then


\(\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}\)

\(\displaystyle \frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}\)
Here's your error. \(\displaystyle \sqrt{3}- i\) has argument \(\displaystyle -\pi/6\) or \(\displaystyle 5\pi/6\), not \(\displaystyle -\pi/12\) or \(\displaystyle 11\pi/12\).
The right triangle formed by the lines from (0, 0) to \(\displaystyle (\sqrt{3}, -1)\), (0, 0) to \(\displaystyle \sqrt{3}, 0\), and from \(\displaystyle (\sqrt{3}, 0)\) to \(\displaystyle (\sqrt{3}, -1)\) is half of an equilateral triangle. It vertex angle is 30 degrees or \(\displaystyle \pi/6\) radians.

anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.?
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
Here's your error. \(\displaystyle \sqrt{3}- i\) has argument \(\displaystyle -\pi/6\) or \(\displaystyle 5\pi/6\), not \(\displaystyle -\pi/12\) or \(\displaystyle 11\pi/12\).
The right triangle formed by the lines from (0, 0) to \(\displaystyle (\sqrt{3}, -1)\), (0, 0) to \(\displaystyle \sqrt{3}, 0\), and from \(\displaystyle (\sqrt{3}, 0)\) to \(\displaystyle (\sqrt{3}, -1)\) is half of an equilateral triangle. It vertex angle is 30 degrees or \(\displaystyle \pi/6\) radians.
And the modulus of \(\displaystyle \sqrt{3} - i\) has been forgotten in the denominator.