Let z be a complex number such that |z| = 1. Find the largest possible value of |z^2 - 1|.
+118703 Let z be a complex number such that |z| = 1. Find the largest possible value of |z^2 - 1|
z=1eiθz2=e2θiz2−1=e2θi−1z2−1=cos(2θ)+isin(2θ)−1z2−1=[cos(2θ)−1]+isin(2θ)|z2−1|2=[cos(2θ)−1]2+sin2(2θ)|z2−1|2=[cos2(2θ)+1−2cos(2θ)]2+sin2(2θ)|z2−1|2=[2−2cos(2θ)]2|z2−1|2=4[1−cos(2θ)]2|z2−1|=2[1−cos(2θ)]
−1≤cos(2θ)≤1−1≤−cos(2θ)≤10≤1−cos(2θ)≤20≤2[1−cos(2θ)]≤40≤|z2−1|≤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\\
