Question and Answers Forum

1) solve x^5 =1 2) if x_i are roots of this equation prove that Σ_i x_i =0 3)prove that cos(((2π)/5)) +cos(((4π)/5)) =−(1/2) 4)calculate cos(((4π)/5)) interms cos(((2π)/5)) then find its values.

$$\left.\mathrm{1}\right)\:{solve}\:{x}^{\mathrm{5}} =\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{if}\:{x}_{{i}} \:{are}\:{roots}\:{of}\:{this}\:{equation}\:{prove}\:{that}\:\sum_{{i}} \:{x}_{{i}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right){prove}\:{that}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\mathrm{4}\right){calculate}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:{interms}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:{then}\:{find}\:{its} \\ $$$${values}. \\ $$

Question Number 31034 Answers: 0 Comments: 0

1) solve inside C z^4 =6 2)sove inside C (((z+i)/(z−i)))^3 +(((z+i)/(z−i)))^2 +(((z+i)/(z−i))) +1=0

$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{4}} =\mathrm{6} \\ $$$$\left.\mathrm{2}\right){sove}\:{inside}\:{C}\:\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{3}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{2}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)\:+\mathrm{1}=\mathrm{0} \\ $$

Question Number 31033 Answers: 0 Comments: 0

factorize p(x)=(1+ix)^n −e^(inθ) wth θ∈R.

$${factorize}\:{p}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} \:−{e}^{{in}\theta} \:\:\:{wth}\:\theta\in{R}. \\ $$

Question Number 31032 Answers: 0 Comments: 0

let give x_0 =0 ,y_0 =1 and {_(y_n =x_(n−1) +y_(n−1) ) ^(x_n =x_(n−1) −y_(n−1) ) for n≥1 let z_n =x_n +i y_n ∀n∈N 1)calculate z_0 ,z_1 and z_2 2)prove that ∀n∈N^ ,n≥1 z_n =(1+i)z_(n−1) find z_n then find the expression of x_n and y_n 3)let put S_n =z_0 +z_1 +....z_n s_n =x_0 +x_1 +...+x_n s_n ^′ =y_0 +y_1 +...+y_n find those sum interms of n.

$${let}\:{give}\:{x}_{\mathrm{0}} =\mathrm{0}\:,{y}_{\mathrm{0}} =\mathrm{1}\:{and}\:\left\{_{{y}_{{n}} ={x}_{{n}−\mathrm{1}} \:+{y}_{{n}−\mathrm{1}} } ^{{x}_{{n}} ={x}_{{n}−\mathrm{1}} \:−{y}_{{n}−\mathrm{1}} } \:\:\:\:\:\:{for}\:{n}\geqslant\mathrm{1}\:{let}\right. \\ $$$${z}_{{n}} ={x}_{{n}} \:+{i}\:{y}_{{n}} \:\:\:\:\:\forall{n}\in{N} \\ $$$$\left.\mathrm{1}\right){calculate}\:{z}_{\mathrm{0}} \:,{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{n}\in{N}^{} ,{n}\geqslant\mathrm{1}\:\:{z}_{{n}} =\left(\mathrm{1}+{i}\right){z}_{{n}−\mathrm{1}} \:{find}\:{z}_{{n}} {then} \\ $$$${find}\:{the}\:{expression}\:{of}\:{x}_{{n}} \:{and}\:{y}_{{n}} \\ $$$$\left.\mathrm{3}\right){let}\:{put}\:{S}_{{n}} ={z}_{\mathrm{0}} \:+{z}_{\mathrm{1}} \:+....{z}_{{n}} \\ $$$${s}_{{n}} ={x}_{\mathrm{0}} \:+{x}_{\mathrm{1}} \:+...+{x}_{{n}} \\ $$$${s}_{{n}} ^{'} ={y}_{\mathrm{0}} \:+{y}_{\mathrm{1}} \:+...+{y}_{{n}} \:\:{find}\:{those}\:{sum}\:{interms}\:{of}\:{n}. \\ $$$$ \\ $$

Question Number 31031 Answers: 0 Comments: 0

1) solve inside C z^(12) =1 and give the solution at form r e^(iθ) 2)calculate 1+u +u^2 +... +u^n then find the solution of z∈C z^8 +z^4 +1=0

$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{12}} =\mathrm{1}\:{and}\:{give}\:{the}\:{solution}\:{at}\:{form} \\ $$$${r}\:{e}^{{i}\theta} \\ $$$$\left.\mathrm{2}\right){calculate}\:\mathrm{1}+{u}\:+{u}^{\mathrm{2}} \:+...\:+{u}^{{n}} \:{then}\:{find}\:{the}\:{solution} \\ $$$${of}\:{z}\in{C}\:\:\:\:\:\:\:{z}^{\mathrm{8}} \:+{z}^{\mathrm{4}} \:+\mathrm{1}=\mathrm{0} \\ $$

Pg 1772 Pg 1773 Pg 1774 Pg 1775 Pg 1776 Pg 1777 Pg 1778 Pg 1779 Pg 1780 Pg 1781

Contact: info@tinkutara.com