Trouble with complex numbers

**maybealways** · May 7th 2012, 12:25 PM

The question is this:

$|(2+i)(cos\frac{\pi}{6}-i sin\frac{\pi}{6})|$

I'm not sure if what I am doing is right...

My next step was this:
$|(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|$

But I'm not sure what to do after. Is it just simple algebra? Because if it is, I did something really weird and didn't get the answer, which is supposed to be $\sqrt5$

Any help would be wonderful! Any working would be great, to see where I went wrong.

**hairymclairy** · May 7th 2012, 01:18 PM

You're doing fine, just multiply out the brackets and write it as $x+yi$ for some values x and y. Then in order to find the absolute value, you simply take the square root of $x^2+y^2$ . If you picture the complex number $x+yi$ on the complex plane, where the x axis corresponds to the real part and the y axis corresponds to the imaginary part, then $|x+yi|$ is simply the hypotenuse of a right angle triangle with horizontal side of length $x$ and vertical side of length $y$ .

**Plato** · May 7th 2012, 01:59 PM

Originally Posted by maybealways

The question is this:
$|(2+i)(cos\frac{\pi}{6}-i sin\frac{\pi}{6})|$
I'm not sure if what I am doing is right...
My next step was this:
$|(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|$

You can really simplify things by noting $|z\cdot w|=|z|\cdot |w|.$
Thus $|(2+i)(\cos\frac{\pi}{6}-i \sin\frac{\pi}{6})|=|(2+i)|\cdot|(\cos\frac{\pi}{6 }-i \sin\frac{\pi}{6})|$

Moreover $|\cos(\theta)\pm i \sin(\theta)|=1,~\forall\theta\in\mathbb{R}~.$

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