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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 +(12−15i)z^2 +(4+4i)z+16+192i=0 Knowing that it has one real root and a purely imaginary root of equal magnitude.
$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{C}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{z}^{\mathrm{4}} +\left(\mathrm{7}−{i}\right){z}^{\mathrm{3}} +\left(\mathrm{12}−\mathrm{15}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{4}{i}\right){z}+\mathrm{16}+\mathrm{192}{i}=\mathrm{0} \\ $$$$\mathrm{Knowing}\:\mathrm{that}\:\mathrm{it}\:\mathrm{has}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{and}\:\mathrm{a}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitude}. \\ $$
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
$$\mathrm{because}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{root}\:\mathrm{is}\:\mathrm{real}\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{the}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{to}\:\mathrm{equal}\:\mathrm{0} \\ $$$$\begin{cases}{{z}^{\mathrm{4}} +\mathrm{7}{z}^{\mathrm{3}} +\mathrm{12}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{16}=\mathrm{0}}\\{\mathrm{i}\left({z}^{\mathrm{3}} +\mathrm{15}{z}^{\mathrm{2}} −\mathrm{4}{z}−\mathrm{192}\right)=\mathrm{0}\:\Rightarrow\:{z}_{\mathrm{1}} =−\mathrm{4}\:\left(\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{192}\right)}\end{cases}\: \\ $$$$\Rightarrow\:{z}_{\mathrm{2}} =−\mathrm{4i}\vee{z}_{\mathrm{2}} =\mathrm{4i} \\ $$$$\mathrm{trying}\:\mathrm{in}\:\mathrm{original}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$${z}_{\mathrm{2}} =\mathrm{4i} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{4}} +\left(\mathrm{7}−\mathrm{i}\right){z}^{\mathrm{3}} +\left(\mathrm{12}−\mathrm{15i}\right){z}^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{4i}\right){z}+\mathrm{16}+\mathrm{192i}= \\ $$$$=\left({z}+\mathrm{4}\right)\left({z}−\mathrm{4i}\right)\left({z}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3i}\right){z}−\mathrm{12}+\mathrm{i}\right) \\ $$$$\Rightarrow \\ $$$${z}_{\mathrm{3}} =−\mathrm{5}−\mathrm{2i} \\ $$$${z}_{\mathrm{4}} =\mathrm{2}−\mathrm{i} \\ $$
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)) :}
$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{formula} \\ $$$${z}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3i}\right){z}−\mathrm{12}+\mathrm{i}=\mathrm{0} \\ $$$${z}=−\frac{\mathrm{3}+\mathrm{3i}}{\mathrm{2}}\pm\sqrt{\frac{\left(\mathrm{3}+\mathrm{3i}\right)^{\mathrm{2}} }{\mathrm{4}}−\left(−\mathrm{12}+\mathrm{i}\right)}= \\ $$$$=−\frac{\mathrm{3}+\mathrm{3i}}{\mathrm{2}}\pm\sqrt{\mathrm{12}+\frac{\mathrm{7}}{\mathrm{2}}\mathrm{i}} \\ $$$$\mathrm{solving}\:\left({a}+{b}\mathrm{i}\right)^{\mathrm{2}} =\mathrm{12}+\frac{\mathrm{7}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:\sqrt{\mathrm{12}+\frac{\mathrm{7}}{\mathrm{2}}\mathrm{i}}=\frac{\mathrm{7}+\mathrm{i}}{\mathrm{2}} \\ $$$${z}=−\frac{\mathrm{3}+\mathrm{3i}}{\mathrm{2}}\pm\frac{\mathrm{7}+\mathrm{i}}{\mathrm{2}}=\begin{cases}{−\mathrm{5}−\mathrm{2i}}\\{\mathrm{2}−\mathrm{i}}\end{cases} \\ $$
Commented by Ar Brandon last updated on 10/Nov/22
Let me try ⇒z^2 +(3+3i)z−12+i=0 ⇒z=((−3−3i±(√((3+3i)^2 +4(12−i))))/2) =((−3−3i±(√(48+14i)))/2)=((−3−3i±(7+i))/2) z_3 =−5−2i , z_4 =2−i
$$\mathrm{Let}\:\mathrm{me}\:\mathrm{try} \\ $$$$\Rightarrow{z}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3}{i}\right){z}−\mathrm{12}+{i}=\mathrm{0} \\ $$$$\Rightarrow{z}=\frac{−\mathrm{3}−\mathrm{3}{i}\pm\sqrt{\left(\mathrm{3}+\mathrm{3}{i}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{12}−{i}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{−\mathrm{3}−\mathrm{3}{i}\pm\sqrt{\mathrm{48}+\mathrm{14}{i}}}{\mathrm{2}}=\frac{−\mathrm{3}−\mathrm{3}{i}\pm\left(\mathrm{7}+{i}\right)}{\mathrm{2}} \\ $$$${z}_{\mathrm{3}} =−\mathrm{5}−\mathrm{2}{i}\:,\:{z}_{\mathrm{4}} =\mathrm{2}−{i} \\ $$
Commented by Ar Brandon last updated on 10/Nov/22
Got it. Thanks Sir !
Commented by Frix last updated on 10/Nov/22
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Commented by Ar Brandon last updated on 10/Nov/22
What about z^2 +(3+3i)z−12+i ? Any quick methode to get it? Appart from Euclid′s division.
$$\mathrm{What}\:\mathrm{about}\:{z}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3}{i}\right){z}−\mathrm{12}+{i}\:? \\ $$$$\mathrm{Any}\:\mathrm{quick}\:\mathrm{methode}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}?\:\mathrm{Appart}\:\mathrm{from} \\ $$$$\mathrm{Euclid}'\mathrm{s}\:\mathrm{division}. \\ $$

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