Qedani Mkhonza
May 24
Use De Moivre's theorem to find z^6 ,and leave your answer in polar form if z=-25 ?
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ChatGPT
May 24
De Moivre's theorem states that for any complex number z = r(cosθ + isinθ), then z^n = r^n(cos(nθ) + isin(nθ).
Given z = -25, we can write it in polar form as z = 25(cosπ + isinπ).
Now, using De Moivre's theorem, we can find z^6:
z^6 = 25^6(cos(6π) + isin(6π))
z^6 = 25^6(cos(0) + isin(0))
z^6 = 25^6(1 + i0)
z^6 = 25^6
Therefore, z^6 = 244140625 in polar form.
Given z = -25, we can write it in polar form as z = 25(cosπ + isinπ).
Now, using De Moivre's theorem, we can find z^6:
z^6 = 25^6(cos(6π) + isin(6π))
z^6 = 25^6(cos(0) + isin(0))
z^6 = 25^6(1 + i0)
z^6 = 25^6
Therefore, z^6 = 244140625 in polar form.
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Google Gemini AI
May 24
[De Moivre's Theorem]
$$z^n=r^n(\cos n\theta +i\sin n\theta)$$
[Given]
$$z=-25=5^2(\cos 270\degree+i\sin 270\degree)$$
[Solution]
$$z^6=5^{12}(\cos 1620\degree+i\sin 1620\degree)$$
$$=5^{12}(\cos (2\times 810\degree )+i\sin( 2\times 810\degree))$$
$$=5^{12}(\cos 0\degree+i\sin 0\degree)$$
$$=5^{12}(1+0i)$$
$$=\boxed{15625}$$
$$z^n=r^n(\cos n\theta +i\sin n\theta)$$
[Given]
$$z=-25=5^2(\cos 270\degree+i\sin 270\degree)$$
[Solution]
$$z^6=5^{12}(\cos 1620\degree+i\sin 1620\degree)$$
$$=5^{12}(\cos (2\times 810\degree )+i\sin( 2\times 810\degree))$$
$$=5^{12}(\cos 0\degree+i\sin 0\degree)$$
$$=5^{12}(1+0i)$$
$$=\boxed{15625}$$
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