Question

Determine z tal que z² = i.

Answer 1

$z=a+bi$
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$z^{2}=i\\(a+bi)^{2}=i\\a^{2}+2abi+(bi)^{2}=i\\a^{2}+2abi+b^{2}i^{2}=i\\a^{2}+2abi+b^{2}(-1)=i\\a^{2}+2abi-b^{2}=i\\(a^{2}-b^{2})+2abi-i=0\\(a^{2}-b^{2})+(2ab-1)i=\\(a^{2}-b^{2})+(2ab-1)i=0+0i$

Igualando as partes reais e as imaginárias:

$a^{2}-b^{2}=0\\a^{2}=b^{2}\\a=\pm\sqrt{b^{2}}\\a=\pm b$

Considerando a = b:

$2ab-1=0\\2b*b-1=0\\2b^{2}-1=0\\2b^{2}=1\\b^{2}=1/2\\b=\pm\sqrt{1/2}\\b=\pm\sqrt{2}/2$

$Como~a=b:\\\\a=\pm\sqrt{2}/2$

$z=(\sqrt{2}/2)+(\sqrt{2}/2)i~~ou~~z=-(\sqrt{2}/2)-(\sqrt{2}/2)i$

Answer 2

(a + bi)² = i

 a² + 2abi + b²i² = i
 a² + 2abi + b²(-1) = i
 a² + 2abi - b² = i

a² - b² + 2abi = i

a² - b² = 0 ==> a² = b² ==> a = √b² ==> a = +/- b
     2ab = 1==> 2a.a = 1 ==> 2a² ==> a² = 1 ==> a = +/- √1 = 1.√2 = √2
                                                                  2                   √2   √2.√2     2
a= +/- √2    e    b = +/- √2 
           2                       2


Z1 =  √2   +  √2 i
           2         2

Z2 = - √2   -   √2 i
           2         2