1+(z+2i)+(z+2i)^2 +(z+2i)^3 +(z+2i)^4 =0 find z , z∈C


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Question Number 70914 by 20190927 last updated on 09/Oct/19

$1 + (z + 2 i) + {(z + 2 i)}^{2} + {(z + 2 i)}^{3} + {(z + 2 i)}^{4} = 0$ $find z, z \in C$
Commented by mathmax by abdo last updated on 09/Oct/19
$let x=z+2i (e) ⇔1+x+x^2 +x^3 +x^4 =0 ⇒((1−x^5 )/(1−x)) =0 and x≠1 ⇒ x^5 =1 and x≠1 butx^5 =1 ⇒x^5 =e^(i2kπ) ⇒x_k =e^((i2kπ)/5) and k∈[[1,4]] so the roots of (e) are z_k =x_k −2i z_k =e^((i2kπ)/5) −2i =cos(((2kπ)/5))+i(sin(((2kπ)/5))−2)=r_k e^(iθ_k ) with r_k =(√(cos(((2kπ)/5))^2 +(sin(((2kπ)/5))−2)^2 )) and θ_k =arctan{((sin(((2kπ)/5))−2)/(cos(((2kπ)/5))))} 1≤k≤4$
$l e t x = z + 2 i (e) \Leftrightarrow 1 + x + x^{2} + x^{3} + x^{4} = 0 \Rightarrow \frac{1 - x^{5}}{1 - x} = 0 a n d x \neq 1 \Rightarrow$ $x^{5} = 1 a n d x \neq 1 {b u t x}^{5} = 1 \Rightarrow x^{5} = e^{i 2 k π} \Rightarrow x_{k} = e^{\frac{i 2 k π}{5}}$ $a n d k \in [[1, 4]] s o t h e r o o t s o f (e) a r e z_{k} = x_{k} - 2 i$ $z_{k} = e^{\frac{i 2 k π}{5}} - 2 i = c o s (\frac{2 k π}{5}) + i (s i n (\frac{2 k π}{5}) - 2) = r_{k} e^{i θ_{k}}$ $w i t h r_{k} = \sqrt{c o s {(\frac{2 k π}{5})}^{2} + {(s i n (\frac{2 k π}{5}) - 2)}^{2}}$ $a n d θ_{k} = a r c t a n {\frac{s i n (\frac{2 k π}{5}) - 2}{c o s (\frac{2 k π}{5})}} 1 ⩽ k ⩽ 4$
Commented by 20190927 last updated on 10/Oct/19

$thank you$
Commented by mathmax by abdo last updated on 10/Oct/19

$y o u a r e w e l c o m e .$
	Answered by mind is power last updated on 09/Oct/19
	$⇒(z+2i −1).(1+(z+2i)+(z+2i)^2 +(z+2i)^3 +(z+2i)^4 )=(z+2i)^5 −1=0 ⇒(z+2i)^5 =e^(2ikπ) ⇒z=e^((2ikπ)/5) −2i k∈{0,....4} k≠0 for our solution$
	$\Rightarrow (z + 2 i - 1) . (1 + (z + 2 i) + {(z + 2 i)}^{2} + {(z + 2 i)}^{3} + {(z + 2 i)}^{4}) = {(z + 2 i)}^{5} - 1 = 0$ $\Rightarrow {(z + 2 i)}^{5} = e^{2 i k π}$ $\Rightarrow z = e^{\frac{2 i k π}{5}} - 2 i k \in {0, . . . . 4}$ $k \neq 0 f o r o u r s o l u t i o n$
	Commented by 20190927 last updated on 10/Oct/19

	$thank you$
	Commented by mind is power last updated on 10/Oct/19

	$y^{'} r e w e l c o m$
	Answered by MJS last updated on 09/Oct/19

	$z = x - 2 i$ $x^{4} + x^{3} + x^{2} + x + 1 = 0$ $(x - 1) (x^{4} + x^{3} + x^{2} + x + 1) = 0$ $x^{5} - 1 = 0$ $\Rightarrow x_{0} = 1 not valid because we added it$ $x_{1} = - \frac{1}{4} + \frac{\sqrt{5}}{4} + \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i$ $x_{2} = - \frac{1}{4} - \frac{\sqrt{5}}{4} + \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i$ $x_{3} = - \frac{1}{4} - \frac{\sqrt{5}}{4} - \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i$ $x_{4} = - \frac{1}{4} + \frac{\sqrt{5}}{4} - \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i$ $z_{i} = x_{i} - 2 i$
	Commented by 20190927 last updated on 10/Oct/19

	$thank you$
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