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\(|z|=1\\ \Re(z)=a\\ \Re(z^2)=\,?\)
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 Jun 11, 2019
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\(\displaystyle \Re{(z)} = a,\text{ so let }z = a +ib.\)

 

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

 

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

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 Jun 12, 2019

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