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Question Number 128080 by mohammad17 last updated on 04/Jan/21

	Answered by mathmax by abdo last updated on 04/Jan/21
	$z^4 −1=(√3)i ⇒z^4 =1+(√3)i =2((1/2)+i((√3)/2))=2e^((iπ)/3) let z =re^(iθ) z^4 =1+(√3)i ⇒r^4 e^(4iθ) =2e^(i(π/3)+2ikπ) ⇒r=^4 (√2) and 4θ =(π/3)+2kπ ⇒ θ=(π/(12))+((kπ)/2) so the roots are z_k =^4 (√2) e^(i((π/(12))+((kπ)/2))) with k∈[[0,3]] z_0 +z_2 =^4 (√2){ e^((iπ)/(12)) +e^((iπ)/(12)) .e^(iπ) } =^4 (√2){e^((iπ)/(12)) −e^((iπ)/(12)) }=0 z_1 +z_3 =^4 (√2){e^((iπ)/(12)) .e^((iπ)/2) +e^((iπ)/(12)) .e^(i((3π)/2)) } =^4 (√2){i e^((iπ)/(12)) −ie^((iπ)/(12)) } =0$
	$z^{4} - 1 = \sqrt{3} i \Rightarrow z^{4} = 1 + \sqrt{3} i = 2 (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = {2 e}^{\frac{i π}{3}} let z = {re}^{i θ}$ $z^{4} = 1 + \sqrt{3} i \Rightarrow r^{4} e^{4 i θ} = {2 e}^{i \frac{π}{3} + 2 ik π} \Rightarrow r =^{4} \sqrt{2} and 4 θ = \frac{π}{3} + 2 k π \Rightarrow$ $θ = \frac{π}{12} + \frac{k π}{2} so the roots are z_{k} =^{4} \sqrt{2} e^{i (\frac{π}{12} + \frac{k π}{2})} with k \in [[0, 3]]$ $z_{0} + z_{2} =^{4} \sqrt{2} {e^{\frac{i π}{12}} + e^{\frac{i π}{12}} . e^{i π}} =^{4} \sqrt{2} {e^{\frac{i π}{12}} - e^{\frac{i π}{12}}} = 0$ $z_{1} + z_{3} =^{4} \sqrt{2} {e^{\frac{i π}{12}} . e^{\frac{i π}{2}} + e^{\frac{i π}{12}} . e^{i \frac{3 π}{2}}} =^{4} \sqrt{2} {i e^{\frac{i π}{12}} - {ie}^{\frac{i π}{12}}} = 0$
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