Problem with complex numbers

**nafix** · February 1st 2008, 03:38 AM

I've been trying to figure out complex numbers as vectors but the book is hard to understand. Can someone please help me out on these problems, my homeworks due in 4 hours.
1.Show which of the following subsets R^3 are subspaces and which are not:
a.The set of vectors in R^3 with first component 1;
b.All vectors x3=0;
c.The set of vectors with nonzero first component.

2.Polar form of complex number:
Write the complex number z=sqrt(3) + i in polar form and find its magnitude and angle in the complex plane.

3.Compute z/zbar where z=3-4i

**topsquark** · February 1st 2008, 07:04 AM

Originally Posted by nafix

2.Polar form of complex number:
Write the complex number z=sqrt(3) + i in polar form and find its magnitude and angle in the complex plane.

$\sqrt{3} + i = re^{i \theta} = r \cdot cos(\theta) + i~r \cdot sin(\theta)$
for some r and $\theta$ .

So we have that
$r \cdot cos(\theta) = \sqrt{3}$
and
$r \cdot sin(\theta) = 1$

Thus
$tan(\theta) = \frac{sin(\theta)}{cos(\theta}) = \frac{r \cdot sin(\theta)}{r \cdot cos(\theta)} = \frac{1}{\sqrt{3}}$

So
$\theta = tan^{-1} \left ( \frac{1}{\sqrt{3}} \right )$

Since both sine and cosine are positive here, $\theta$ is a QI angle. Thus
$\theta = tan^{-1} \left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi}{6}$

To find r note that
$r = \sqrt{( r \cdot cos(\theta) )^2 + ( r \cdot sin(\theta) )^2} = \sqrt{(\sqrt{3})^2 + (1)^2} = 2$

So
$\sqrt{3} + i = 2e^{i \pi/6}$

-Dan

**topsquark** · February 1st 2008, 07:11 AM

Originally Posted by nafix

3.Compute z/zbar where z=3-4i

$\bar{z} = 3 + 4i$

So
$\frac{z}{\bar{z}} = \frac{3 - 4i}{3 + 4i}$

Now to rationalize the denominator:
$= \frac{3 - 4i}{3 + 4i} = \frac{3 - 4i}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} = \frac{(3 - 4i)^2}{(3 + 4i)(3 - 4i)}$

$= \frac{9 - 24i + 16i^2}{9 - 16i^2}$

$= \frac{9 - 24i - 16}{9 + 16}$

$= \frac{-7 - 24i}{25}$

So if you like
$\frac{3 - 4i}{3 + 4i} = -\frac{7}{25} - i~\frac{24}{25}$

-Dan

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