# Thread: Simplifying an expression involving complex numbers.

1. ## Simplifying an expression involving complex numbers.

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.?

2. Originally Posted by Henryt999
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.?

3. Originally Posted by HallsofIvy
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.