Solve the equation: (1) z^3 + 1 − 10i = 0 (2) z^4 − i + 2 = 0


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Question Number 51680 by Tawa1 last updated on 29/Dec/18

$Solve the equation :$ $(1) z^{3} + 1 - 10 i = 0$ $(2) z^{4} - i + 2 = 0$
Commented by maxmathsup by imad last updated on 29/Dec/18

$2) z^{4} - i + 2 = 0 \Leftrightarrow z^{4} = - 2 + i = \sqrt{5} (- \frac{2}{\sqrt{5}} + \frac{i}{\sqrt{5}}) = r (c o s θ + i s i n θ) \Rightarrow$ $r = \sqrt{5} a n d c o s θ = - \frac{2}{\sqrt{5}} a n d s i n θ = \frac{1}{\sqrt{5}} \Rightarrow t a n θ = - \frac{1}{2} \Rightarrow θ = - a r c t a n (\frac{1}{2}) \Rightarrow$ $- 2 + i = r e^{- i a r c t a n (\frac{1}{2})} l e t z = ρ e^{i α} s o z^{4} = - 2 + i \Rightarrow ρ^{4} = r a n d 4 α = - a r c t a n (\frac{1}{2}) + 2 k π \Rightarrow$ $ρ =^{4} \sqrt{r} a n d α_{k} = - \frac{1}{4} a r c t a n (\frac{1}{2}) + \frac{k π}{2} w i t h k \in [[0, 3]] s o t h e r o o t s o f t h i s$ $e q u a t i o n a r e Z_{k} =^{8} \sqrt{5} e^{i (- \frac{1}{4} a r c t a n (\frac{1}{2}) + \frac{k π}{2})} a n d k \in [[0, 3]] .$
Commented by maxmathsup by imad last updated on 29/Dec/18

$f o r q u e s t i o n 1) w e f o l l o w t h e s a m e m e t h o d .$
	Answered by peter frank last updated on 29/Dec/18

	$1) z^{3} = - 1 + 10 i$ $r = \sqrt{101}$ $θ = \tan^{- 1} - \frac{10}{1} = - 84 . 2894^{°}$ $. . . . . .$ $2) z^{4} = - 2 + i$ $r = \sqrt{5}$ $θ = \tan^{- 1} \frac{- 2}{1} = - 63.4349$ $. . . . . p l e a s e w a i t . . .$
	Commented by Tawa1 last updated on 29/Dec/18

	$Please is this the solution ? ? ?$
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