Trouble with complex numbers

The question is this:

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

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

My next step was this:

$\displaystyle |(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|$

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 $\displaystyle \sqrt5$

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

Re: Trouble with complex numbers

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

Re: Trouble with complex numbers

Quote:

Originally Posted by

**maybealways** The question is this:

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

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

My next step was this:

$\displaystyle |(2+i)(\frac{\sqrt3}{2}-\frac{i}{2})|$

You can really simplify things by noting $\displaystyle |z\cdot w|=|z|\cdot |w|.$

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

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