1:st year University math

Hello

(This may be placed in the wrong part of this forum, so please move it if thats the case.)

I have just start to read math analysis 1 and i have some problems to understand, so i ask you for some help.

All solution to:

(1 + i)z^{2} − (6 + 4i)z + 9 + 7i = 0,

All solutions to:

(z + 1 + 2i)^{3} = 27i

If someone can show me how to solve this, i hope i can solve my other problems :)

Thanks

Re: 1:st year University math

Quote:

Originally Posted by

**spope** Hello

(This may be placed in the wrong part of this forum, so please move it if thats the case.)

I have just start to read math analysis 1 and i have some problems to understand, so i ask you for some help.

All solution to:

(1 + i)z^{2} − (6 + 4i)z + 9 + 7i = 0,

All solutions to:

(z + 1 + 2i)^{3} = 27i

If someone can show me how to solve this, i hope i can solve my other problems :)

Thanks

The first is a quadratic, so use the quadratic formula.

For the second take the cube root of both sides. What is the cube root of i? (Hint: There are three of them. Can you put them into a simple form?)

-Dan

Re: 1:st year University math

The first one i just replace z with a+ib and then multiplice?

And the second is (z+1+2i)=3i?

Re: 1:st year University math

Quote:

Originally Posted by

**spope** The first one i just replace z with a+ib and then multiplice?

And the second is (z+1+2i)=3i?

Yeah, it probably would be easier to attack the first problem that way. :)

For the second problem, the cube root of i is not i. Rewrite $\displaystyle i = e^{i \pi/2 + 2ni \pi}$, where n is an integer. What is the cube root of this?

-Dan

Re: 1:st year University math

Quote:

Originally Posted by

**topsquark** Yeah, it probably would be easier to attack the first problem that way. :)

For the second problem, the cube root of i is not i. Rewrite $\displaystyle i = e^{i \pi/2 + 2ni \pi}$, where n is an integer. What is the cube root of this?

-Dan

So $\displaystyle (z+1+2i)^{3}= 27 e^{i \pi/2 + 2ni \pi}$?

If it was $\displaystyle z^{3}= 27 e^{i \pi/2 + 2ni \pi}$ i can solve it. But what happen if it is $\displaystyle (z+1+2i)^{3}$?

Re: 1:st year University math

Hi again

Thanks for the help topsquark :) It was pretty easy when i know how to do(Itwasntme)

I have some problem solving this one now.

http://puu.sh/5ZmPT.jpg

Any idea how to solve from here?

Re: 1:st year University math

Quote:

Originally Posted by

**spope** Hi again

Thanks for the help topsquark :) It was pretty easy when i know how to do(Itwasntme)

I have some problem solving this one now.

http://puu.sh/5ZmPT.jpg
Any idea how to solve from here?

I presume the exponent "3" in the first line is a typo?

Let $\displaystyle y = cos( \theta )$. This turns your equation into a quadratic.

-Dan

Re: 1:st year University math

Quote:

Originally Posted by

**topsquark** I presume the exponent "3" in the first line is a typo?

Let $\displaystyle y = cos( \theta )$. This turns your equation into a quadratic.

-Dan

You are right. It should be $\displaystyle sin^2(x)-sqrt(3)cos(x)=7/4$

I have used $\displaystyle \sin^2 \theta + \cos^2 \theta = 1$ to switch from $\displaystyle sin^2(x)$ to $\displaystyle 1-cos^2(x)$

$\displaystyle y = cos( \theta )$? i don't know how to proceed from there.