# Simplifying an expression involving complex numbers.

#### 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

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

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

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

#### HallsofIvy

MHF Helper
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
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.