Math Help - Hi I have 4 questions I need help with

1. Hi I have 4 questions I need help with

Any or all would be helpful

r - 5 sin(theta) = 5 cos (theta)

to rectangular form

------------

x^2 + 1 - (square root of 3i)

find all solutions, imaginary too.

-----------

Find solutions for 5th root of 1

I don't even get this at all???

-----------

Polar coordinates of (square root of 2), (square root of -2)

-----------

Reminder I needed to do this without calculator. There were the hardest questions on my homework I did not get!!! And my teacher is a jerk and he always collects our homework but I don't get this!!! I am getting a tutor because he teaches me NOTHING.

Thanks,

Kyle

2. For some of these all you need to do is apply the definitions.

Originally Posted by DoubleX
r - 5 sin(theta) = 5 cos (theta)

to rectangular form
$x = r~cos(\theta)$
$y = r~sin(\theta)$

So what would this be?

x^2 + 1 - (square root of 3i)

find all solutions, imaginary too.
This isn't an equation, it's an expression. For this to be an equation there needs to be an equal sign. Is it supposed to be
$x^2 + 1 - \sqrt{3i} = 0$

Find solutions for 5th root of 1
Convert the number 1 to the polar form for a complex number.[/quote]
$\sqrt[5]{1} = (1)^{1/5} = (re^{i \theta})^{1/5}$
where
$re^{i \theta} = 1$

Thus r = 1 and $\theta = 0$. However, note that we can always multiply $re^{i \theta}$ by a $e^{2ni \pi} = 1$ where n is an integer. So
$\sqrt[5]{1} = (1 \cdot e^{2ni \pi})^{1/5}$

$= e^{2ni \pi / 5}$

Now put n = 0, 1, 2, 3, and 4.

Polar coordinates of (square root of 2), (square root of -2)
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$

So we know that
$x^2 + y^2 = r^2~cos^2(\theta) + r^2~sin^2(\theta) = r^2(cos^2(\theta) + sin^2(\theta)) = r^2$

So $r = \sqrt{x^2 + y^2}$

Again
$\frac{y}{x} = \frac{r~sin(\theta)}{r~cos(\theta)} = tan(\theta)$

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

-Dan

3. For some of these all you need to do is apply the definitions.

$x = r~cos(\theta)$
$y = r~sin(\theta)$

So what would this be?

_______
maybe
r-y=x
r=5
???

-----------------

This isn't an equation, it's an expression. For this to be an equation there needs to be an equal sign. Is it supposed to be
$x^2 + 1 - \sqrt{3i} = 0$

_____

thats what i meant sorry

-------------------
Convert the number 1 to the polar form for a complex number.[/quote]
$\sqrt[5]{1} = (1)^{1/5} = (re^{i \theta})^{1/5}$
where
$re^{i \theta} = 1$

_____________

Thus r = 1 and $\theta = 0$. However, note that we can always multiply $re^{i \theta}$ by a $e^{2ni \pi} = 1$ where n is an integer. So
$\sqrt[5]{1} = (1 \cdot e^{2ni \pi})^{1/5}$

$= e^{2ni \pi / 5}$

Now put n = 0, 1, 2, 3, and 4.

_______________-

we have a different formula I think, where its a=( $\theta$ + 360 degrees * k)/n

I still don't understand it though.

----------------------------
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$

So we know that
$x^2 + y^2 = r^2~cos^2(\theta) + r^2~sin^2(\theta) = r^2(cos^2(\theta) + sin^2(\theta)) = r^2$

So $r = \sqrt{x^2 + y^2}$

Again
$\frac{y}{x} = \frac{r~sin(\theta)}{r~cos(\theta)} = tan(\theta)$

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

____________

Thanks, but I know that and I have to find it without a calculator so I don't know how.
---------
-Dan

thank you!

4. Originally Posted by DoubleX
For some of these all you need to do is apply the definitions.

$x = r~cos(\theta)$
$y = r~sin(\theta)$

So what would this be?

_______
maybe
r-y=x
r=5
???
Multiply both sides of your original equation by r. Then you have
$r^2 - r~sin(\theta) = r~cos(\theta)$

Now, what is $r^2$ in terms of x and y?

-Dan

5. Originally Posted by DoubleX
$x^2 + 1 - \sqrt{3i} = 0$
One more question: At this level shouldn't the equation be $x^2 + 1 - \sqrt{3} \cdot i =$?

$x^2 + 1 - \sqrt{3i} = 0$

$x^2 = -1 + \sqrt{3i}$

I'd put the right hand side into polar form at this point. It would be best to start with the
$\sqrt{3i} = (3i)^{1/2} = \sqrt{3} \cdot i^{1/2} = \sqrt{3} ( e^{i \pi} )^{1/2}$

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

$= \sqrt{3} \left ( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right )$

$= \frac{\sqrt{6}}{2} + i \frac{\sqrt{6}}{2}$

So
$-1 + \sqrt{3i} = \left ( -1 + \frac{\sqrt{6}}{2} \right ) + i \frac{\sqrt{6}}{2}$

Then you find the polar form of that and take the square root of it.

This is just too messy for someone working at the level you seem to be. Please confirm that I have the right equation and I'll continue if you need it.

-Dan

6. Originally Posted by DoubleX
Convert the number 1 to the polar form for a complex number.
$\sqrt[5]{1} = (1)^{1/5} = (re^{i \theta})^{1/5}$
where
$re^{i \theta} = 1$

_____________

Thus r = 1 and $\theta = 0$. However, note that we can always multiply $re^{i \theta}$ by a $e^{2ni \pi} = 1$ where n is an integer. So
$\sqrt[5]{1} = (1 \cdot e^{2ni \pi})^{1/5}$

$= e^{2ni \pi / 5}$

Now put n = 0, 1, 2, 3, and 4.

_______________-

we have a different formula I think, where its a=( $\theta$ + 360 degrees * k)/n

I still don't understand it though.
[/quote]
The only difference is I'm expressing the polar angle in radians rather than degrees. So
$\sqrt[5]{1} = (1 \cdot e^{360ni})^{1/5}$

$= e^{360ni/5} = e^{72ni}$

Now put n = 0, 1, 2, 3, and 4.

-Dan

7. Originally Posted by DoubleX
$x = r~cos(\theta)$
and
$y = r~sin(\theta)$

So we know that
$x^2 + y^2 = r^2~cos^2(\theta) + r^2~sin^2(\theta) = r^2(cos^2(\theta) + sin^2(\theta)) = r^2$

So $r = \sqrt{x^2 + y^2}$

Again
$\frac{y}{x} = \frac{r~sin(\theta)}{r~cos(\theta)} = tan(\theta)$

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

____________

Thanks, but I know that and I have to find it without a calculator so I don't know how.
Have you tried the hint:
$(x, y) = ( \sqrt{2}, -\sqrt{2} )$

$r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} =$?

$\theta = tan^{-1} \left ( \frac{y}{x} \right ) = tan^{-1} \left ( \frac{-\sqrt{2}}{\sqrt{2}} \right ) = tan^{-1}(-1)$

Surely you can do these without a calculator!

-Dan

8. Thank you very very much topsquark!!! Thats enough for me to work with

EDIT: except one more thing, in the fifth root of 1, in our formula a=(theta+360degrees * k)/n, how do we know what theta is? Thank you.

9. Originally Posted by DoubleX
Thank you very very much topsquark!!! Thats enough for me to work with

EDIT: except one more thing, in the fifth root of 1, in our formula a=(theta+360degrees * k)/n, how do we know what theta is? Thank you.
In that particular case $\theta = 0$ as we were representing the number 1. I left it general to (hopefully) make things a bit clearer about the general approach to this problem.

-Dan