Simplifying an expression involving complex numbers.

**Henryt999** · May 20th 2010, 04:12 AM

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

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

and that becomes:

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

and then

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

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

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

**HallsofIvy** · May 20th 2010, 04:25 AM

Originally Posted by Henryt999

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

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

and that becomes:

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

and then

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

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

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

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

**mr fantastic** · May 20th 2010, 04:31 AM

Originally Posted by HallsofIvy

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

And the modulus of $\sqrt{3} - i$ has been forgotten in the denominator.

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