Question

Să se calculeze modul nr complex z=(3+i)^4.Dau coroană repede!​

Answer 1

$\displaystyle\it\\modulul~unui~numar~complex~z=a+bi,~este:\boxed{\it|z|=\sqrt{a^2+b^2}}~.\\\boxed{\it i=\sqrt{-1}}~.\\------------------------------\\z=(3+i)^4=(3+i)^2(3+i)^2=(9+6i+i^2)(9+6i+i^2)=\\(9+6i-1)(9+6i-1)=(8+6i)(8+6i)=28+96i.\\|z|=\sqrt{\underbrace{28^2}_{a}+\underbrace{96^2}_{b}}=\boxed{\it 100}.$