# 1:st year University math

• November 26th 2013, 11:33 AM
spope
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)z2 − (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
• November 26th 2013, 11:40 AM
topsquark
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)z2 − (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

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
• November 26th 2013, 11:44 AM
spope
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?
• November 26th 2013, 11:56 AM
topsquark
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 $i = e^{i \pi/2 + 2ni \pi}$, where n is an integer. What is the cube root of this?

-Dan
• November 28th 2013, 01:03 AM
spope
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 $i = e^{i \pi/2 + 2ni \pi}$, where n is an integer. What is the cube root of this?

-Dan

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

If it was $z^{3}= 27 e^{i \pi/2 + 2ni \pi}$ i can solve it. But what happen if it is $(z+1+2i)^{3}$?
• December 27th 2013, 03:45 AM
spope
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?
• December 27th 2013, 12:01 PM
topsquark
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 $y = cos( \theta )$. This turns your equation into a quadratic.

-Dan
• December 27th 2013, 01:13 PM
spope
Re: 1:st year University math
Quote:

Originally Posted by topsquark
I presume the exponent "3" in the first line is a typo?

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

-Dan

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

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

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