Solve in C the equation z^4 +(7−i)z^3 +(12−15i)z^2 +(4+4i)z+16+192i=0 Knowing that it has one real root and a purely imaginary root of equal magnitude.


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Question Number 180274 by Ar Brandon last updated on 09/Nov/22

$Solve in C the equation$ $z^{4} + (7 - i) z^{3} + (12 - 15 i) z^{2} + (4 + 4 i) z + 16 + 192 i = 0$ $Knowing that it has one real root and a purely imaginary root$ $of equal magnitude .$
	Answered by Frix last updated on 09/Nov/22
	$because at least one root is real we need the imaginary part to equal 0 { ((z^4 +7z^3 +12z^2 +4z+16=0)),((i(z^3 +15z^2 −4z−192)=0 ⇒ z_1 =−4 (trying factors of 192))) :} ⇒ z_2 =−4i∨z_2 =4i trying in original equation we get z_2 =4i ⇒ z^4 +(7−i)z^3 +(12−15i)z^2 +(4+4i)z+16+192i= =(z+4)(z−4i)(z^2 +(3+3i)z−12+i) ⇒ z_3 =−5−2i z_4 =2−i$
	$because at least one root is real we need$ $the imaginary part to equal 0$ ${\begin{cases} z^{4} + 7 z^{3} + 12 z^{2} + 4 z + 16 = 0 \\ i (z^{3} + 15 z^{2} - 4 z - 192) = 0 \Rightarrow z_{1} = - 4 (trying factors of 192) \end{cases}$ $\Rightarrow z_{2} = - 4 i \lor z_{2} = 4 i$ $trying in original equation we get$ $z_{2} = 4 i$ $\Rightarrow$ $z^{4} + (7 - i) z^{3} + (12 - 15 i) z^{2} + (4 + 4 i) z + 16 + 192 i =$ $= (z + 4) (z - 4 i) (z^{2} + (3 + 3 i) z - 12 + i)$ $\Rightarrow$ $z_{3} = - 5 - 2 i$ $z_{4} = 2 - i$
	Commented by Ar Brandon last updated on 10/Nov/22
	Thanks. How did you get z3 and Z4 please?
	Commented by Frix last updated on 10/Nov/22
	$you can use the usual formula z^2 +(3+3i)z−12+i=0 z=−((3+3i)/2)±(√((((3+3i)^2 )/4)−(−12+i)))= =−((3+3i)/2)±(√(12+(7/2)i)) solving (a+bi)^2 =12+(7/2)i ⇒ (√(12+(7/2)i))=((7+i)/2) z=−((3+3i)/2)±((7+i)/2)= { ((−5−2i)),((2−i)) :}$
	$you can use the usual formula$ $z^{2} + (3 + 3 i) z - 12 + i = 0$ $z = - \frac{3 + 3 i}{2} \pm \sqrt{\frac{{(3 + 3 i)}^{2}}{4} - (- 12 + i)} =$ $= - \frac{3 + 3 i}{2} \pm \sqrt{12 + \frac{7}{2} i}$ $solving {(a + b i)}^{2} = 12 + \frac{7}{2} i \Rightarrow \sqrt{12 + \frac{7}{2} i} = \frac{7 + i}{2}$ $z = - \frac{3 + 3 i}{2} \pm \frac{7 + i}{2} = {\begin{cases} - 5 - 2 i \\ 2 - i \end{cases}$
	Commented by Ar Brandon last updated on 10/Nov/22

	$Let me try$ $\Rightarrow z^{2} + (3 + 3 i) z - 12 + i = 0$ $\Rightarrow z = \frac{- 3 - 3 i \pm \sqrt{{(3 + 3 i)}^{2} + 4 (12 - i)}}{2}$ $= \frac{- 3 - 3 i \pm \sqrt{48 + 14 i}}{2} = \frac{- 3 - 3 i \pm (7 + i)}{2}$ $z_{3} = - 5 - 2 i, z_{4} = 2 - i$
	Commented by Ar Brandon last updated on 10/Nov/22
	Got it. Thanks Sir !
	Commented by Frix last updated on 10/Nov/22
	��
	Commented by Ar Brandon last updated on 10/Nov/22

	$What about z^{2} + (3 + 3 i) z - 12 + i ?$ $Any quick methode to get it ? Appart from$ ${Euclid}^{'} s division .$
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