# Problem with complex numbers

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• Feb 1st 2008, 04:38 AM
nafix
Problem with complex numbers
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
• Feb 1st 2008, 08:04 AM
topsquark
Quote:

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
• Feb 1st 2008, 08:11 AM
topsquark
Quote:

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