Melody

Thanks Ginger,

I still have trouble getting my head around your method. I don't like working with formulas unless I develop them myself.

This is just becasue I have a bad memory and don't trust myself to apply other people's logic properly.

I usually just prefer to think of all cards or what ever being unique and work back from there.

Let see if i can do it again now, without looking back, and come up with the same answer.

There are 6! ways to line up the cards,  but it doesn't matter what order the pairs are in

so that makes 6!/(2*2*2)= 90 ways

There are 3!*3! ways that they can get each get a red and a yellow = 6*6=36

36/90 = 40%

Great, now we are all in total agreement.  :)    (I hope guest actually learns something.)

There are two standard wasy to do this. 

the GP way is not the easiest but if that is what you think is wanted then lets take a look.

\(0.23\bar{72}\\ =0.23 + 0.0072+0.000072 + \dots\\ consider \\ 0.0072+0.000072 + \dots\\ \text{This is a GP}\\ a=0.0072,\quad r=0.01\\ S_\infty=\frac{a}{1-r}\\ S_\infty=\frac{0.0072}{0.99}\\ S_\infty=\frac{72}{9900}=\frac{8}{1100}\\~\\ 0.23\bar{72}=\frac{23}{100}+\frac{8}{1100}=\frac{253+8}{1100}=\frac{261}{1100} \)

You need to check my working.

I just started doing this when the pic disappeared.  Maybe it was copyrighted??

What are the end points for the linear segments.  

Does 1 look right?   

is 2 the same as 1?

is 3 the same as 2?

is 4 the same as 3?

Where does it go different?

You do not need help with this. There is a calculator on this site if you need to use it.

1)   The first one, start by adding m to BOTH sides  then come back with whtat you have if you need more hgelp

2)   Multiply both sides out and simpleify the two sides seperately first.

remember that   

7-5(3-2x)

= 7+     -5(3-2x)

the minus sign belongs to the 5.

After you do the first bit come back with what you have done.

You have to think of a logical method that might work.

Suggestion

Hoe many ways if only one letter is used

How many ways if 2 letters are used

How many ways if 3 letters are used

How many ways if 4 letters are used.

Thinks about that and come back with some of your own input.

\(\frac{(1 + i\sqrt3)^6}{(1 + i\sqrt3)^4}\\ =(1 + i\sqrt3)^2\)

You can finish it  :)

Let z be a complex number such that |z| = 1. Find the largest possible value of |z^2 - 1|

\(z=1e^{i\theta}\\ z^2=e^{2\theta i}\\ z^2-1=e^{2\theta i}-1\\ z^2-1=cos(2\theta)+isin(2\theta)-1\\ z^2-1=[cos(2\theta)-1]+isin(2\theta)\\ |z^2-1|^2=[cos(2\theta)-1]^2+sin^2(2\theta)\\ |z^2-1|^2=[cos^2(2\theta)+1-2cos(2\theta)]^2+sin^2(2\theta)\\ |z^2-1|^2=[2-2cos(2\theta)]^2\\ |z^2-1|^2=4[1-cos(2\theta)]^2\\ |z^2-1|=2[1-cos(2\theta)]\\\)

\(-1\le cos(2\theta)\le1\\ -1\le -cos(2\theta)\le1\\ 0\le 1-cos(2\theta)\le 2\\ 0\le 2[1-cos(2\theta)]\le 4\\ 0\le |z^2-1|\le 4\\ \)

So the largest possible value is 4 

You need to check what i have done.

LaTex

z=1e^{i\theta}\\
z^2=e^{2\theta i}\\
z^2-1=e^{2\theta i}-1\\
z^2-1=cos(2\theta)+isin(2\theta)-1\\
z^2-1=[cos(2\theta)-1]+isin(2\theta)\\
|z^2-1|^2=[cos(2\theta)-1]^2+sin^2(2\theta)\\
|z^2-1|^2=[cos^2(2\theta)+1-2cos(2\theta)]^2+sin^2(2\theta)\\
|z^2-1|^2=[2-2cos(2\theta)]^2\\
|z^2-1|^2=4[1-cos(2\theta)]^2\\
|z^2-1|=2[1-cos(2\theta)]\\

-1\le cos(2\theta)\le1\\
-1\le -cos(2\theta)\le1\\
0\le 1-cos(2\theta)\le 2\\
0\le 2[1-cos(2\theta)]\le 4\\
0\le |z^2-1|\le 4\\

Thanks Ginger,

That solution does look more convincing   

I understand my old logic but I do not understand what you have done.

Would you care to explain?

Thanks Tiggsy

I can follow all that.  Similar to mine just better :))

\(\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{24 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25 \)

LaTex

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)}
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)}
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{24
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25