se |4+k.i| = 2√5 então
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|4+k*i| = 2sqrt(5)
seria isso ?
seria isso ?
lucia123456789:
as respostas que tem é a) k = +1 b) k+2 c) k+5
sim
Respondido por
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Bom dia Lucia!
Solução!
Modulo de Z
![|z|= \sqrt{4+ki} \\\\\
\sqrt{4+ki}= 2\sqrt{5}\\\\\
i=1\\\\\\
\sqrt{4+ki}=2 \sqrt{5}\\\\\
(\sqrt{4^{2} +k^{2} })^{2} =(2 \sqrt{5})^{2} \\\\\
(\sqrt{16 +k^{2} })^{2} = (2 \sqrt{5})^{2}\\\\\
16+k^{2}=4.5\\\\\
16+k^{2} =20\\\\\ k^{2}=20-16\\\\\ k^{2}=4\\\\\
k= \sqrt{4} \\\\\
k=2\\\\\
\boxed{k=2}](https://tex.z-dn.net/?f=%7Cz%7C%3D+%5Csqrt%7B4%2Bki%7D+%5C%5C%5C%5C%5C%0A+%5Csqrt%7B4%2Bki%7D%3D+2%5Csqrt%7B5%7D%5C%5C%5C%5C%5C%0Ai%3D1%5C%5C%5C%5C%5C%5C%0A+%5Csqrt%7B4%2Bki%7D%3D2+%5Csqrt%7B5%7D%5C%5C%5C%5C%5C+%0A%0A+%28%5Csqrt%7B4%5E%7B2%7D++%2Bk%5E%7B2%7D+%7D%29%5E%7B2%7D+%3D%282+%5Csqrt%7B5%7D%29%5E%7B2%7D+%5C%5C%5C%5C%5C%0A%28%5Csqrt%7B16+%2Bk%5E%7B2%7D+%7D%29%5E%7B2%7D+%3D+%282+%5Csqrt%7B5%7D%29%5E%7B2%7D%5C%5C%5C%5C%5C%0A16%2Bk%5E%7B2%7D%3D4.5%5C%5C%5C%5C%5C%0A16%2Bk%5E%7B2%7D+%3D20%5C%5C%5C%5C%5C+k%5E%7B2%7D%3D20-16%5C%5C%5C%5C%5C+k%5E%7B2%7D%3D4%5C%5C%5C%5C%5C%0Ak%3D+%5Csqrt%7B4%7D+%5C%5C%5C%5C%5C%0Ak%3D2%5C%5C%5C%5C%5C%0A%5Cboxed%7Bk%3D2%7D++%0A+++ "|z|= \sqrt{4+ki} \\\\\
\sqrt{4+ki}= 2\sqrt{5}\\\\\
i=1\\\\\\
\sqrt{4+ki}=2 \sqrt{5}\\\\\
(\sqrt{4^{2} +k^{2} })^{2} =(2 \sqrt{5})^{2} \\\\\
(\sqrt{16 +k^{2} })^{2} = (2 \sqrt{5})^{2}\\\\\
16+k^{2}=4.5\\\\\
16+k^{2} =20\\\\\ k^{2}=20-16\\\\\ k^{2}=4\\\\\
k= \sqrt{4} \\\\\
k=2\\\\\
\boxed{k=2}")
Bom dia!
Bons estudos!
Solução!
Modulo de Z
Bom dia!
Bons estudos!
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