help complex numbers

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\\