+0  
 
0
101
1
avatar

\(|z|=1\\ \Re(z)=a\\ \Re(z^2)=\,?\)
Thanks!

 Jun 11, 2019
 #1
avatar
0

\(\displaystyle \Re{(z)} = a,\text{ so let }z = a +ib.\)

 

\(\displaystyle |z| =\sqrt{ a^{2}+b^{2}}=1, \\ \text{ so }a^{2}+b^{2}=1,\\ b^{2}=1-a^{2}.\)

 

\(\displaystyle z^{2}=(a+ib)(a+ib)=a^{2}-b^{2}+2iab,\\ \text{so }\Re{(z^{2})}=a^{2}-b^{2}=a^{2}-(1-a^{2})=2a^{2}-1.\)

.
 Jun 12, 2019

29 Online Users

avatar
avatar