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Find-the-square-root-of-5-12i-hence-solve-z-2-4-i-z-5-6i-0-

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Question Number 47433 by Tawa1 last updated on 09/Nov/18
Find the square root of − 5 − 12i, hence solve: z^2 − (4 + i)z + (5 + 6i) = 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}\:\:−\:\mathrm{5}\:−\:\mathrm{12i},\:\:\mathrm{hence}\:\mathrm{solve}:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{4}\:+\:\mathrm{i}\right)\mathrm{z}\:+\:\left(\mathrm{5}\:+\:\mathrm{6i}\right)\:=\:\mathrm{0} \\ $$
Commented by peter frank last updated on 10/Nov/18
let a+ib=(√(-5−12i)) a^2 −b^2 +2abi=-5−12i by comparison a^2 −b^2 =-5......(1) 2abi=-12i⇒2ab=-12....(2) a=(±2 or±3) b=(±3 or±2) a+ib=±(2+3i) or±(3+2i)
$$\mathrm{let}\:\mathrm{a}+\mathrm{ib}=\sqrt{-\mathrm{5}−\mathrm{12i}} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{2abi}=-\mathrm{5}−\mathrm{12i} \\ $$$$\mathrm{by}\:\mathrm{comparison} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} =-\mathrm{5}……\left(\mathrm{1}\right) \\ $$$$\mathrm{2abi}=-\mathrm{12i}\Rightarrow\mathrm{2ab}=-\mathrm{12}….\left(\mathrm{2}\right) \\ $$$$\mathrm{a}=\left(\pm\mathrm{2}\:\mathrm{or}\pm\mathrm{3}\right) \\ $$$$\mathrm{b}=\left(\pm\mathrm{3}\:\mathrm{or}\pm\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\mathrm{a}+\mathrm{ib}=\pm\left(\mathrm{2}+\mathrm{3i}\right)\:\mathrm{or}\pm\left(\mathrm{3}+\mathrm{2i}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Nov/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 10/Nov/18
∣−5−12i∣=(√(25+144))=(√(169))=13 ⇒−5−12i =13(−(5/(13)) −((12)/(13))i)=r e^(iπ) ((5/(13))+((12)/(13))i)and (5/(13))+((12)/(13))i=e^(iθ) wih r=13 and cosθ =−(5/(13)) and sinθ =−((12)/(13)) ⇒tanθ=((12)/5) and r=13 and θ=arctan(((12)/(5 ))) ⇒−5−12i =13e^(i(π +arctan(((12)/5))) ⇒ (√(−5−12i))=+^− (√(13))e^(i((π/2)+(1/2)arctan(((12)/5)))) for the equation z^2 −(4+i)z+5+6i =0 Δ =(4+i)^2 −4(5+6i) =16+8i−1−20−24i =−5+16i ⇒ z_1 =((4+i+(√(−5+16i)))/2) and z_2 =((4+i−(√(−5+16i)))/2) we see there no relation between the 2 Questions..
$$\mid−\mathrm{5}−\mathrm{12}{i}\mid=\sqrt{\mathrm{25}+\mathrm{144}}=\sqrt{\mathrm{169}}=\mathrm{13}\:\Rightarrow−\mathrm{5}−\mathrm{12}{i}\:=\mathrm{13}\left(−\frac{\mathrm{5}}{\mathrm{13}}\:−\frac{\mathrm{12}}{\mathrm{13}}{i}\right)={r}\:{e}^{{i}\pi} \left(\frac{\mathrm{5}}{\mathrm{13}}+\frac{\mathrm{12}}{\mathrm{13}}{i}\right){and} \\ $$$$\frac{\mathrm{5}}{\mathrm{13}}+\frac{\mathrm{12}}{\mathrm{13}}{i}={e}^{{i}\theta} \:\:\:{wih}\:{r}=\mathrm{13}\:{and}\:{cos}\theta\:=−\frac{\mathrm{5}}{\mathrm{13}}\:\:{and}\:{sin}\theta\:=−\frac{\mathrm{12}}{\mathrm{13}}\:\Rightarrow{tan}\theta=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$${and}\:{r}=\mathrm{13}\:{and}\:\theta={arctan}\left(\frac{\mathrm{12}}{\mathrm{5}\:}\right)\:\Rightarrow−\mathrm{5}−\mathrm{12}{i}\:=\mathrm{13}{e}^{{i}\left(\pi\:+{arctan}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)\right.} \:\Rightarrow \\ $$$$\sqrt{−\mathrm{5}−\mathrm{12}{i}}=\overset{−} {+}\sqrt{\mathrm{13}}{e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)\right)} \\ $$$${for}\:{the}\:{equation}\:{z}^{\mathrm{2}} −\left(\mathrm{4}+{i}\right){z}+\mathrm{5}+\mathrm{6}{i}\:=\mathrm{0} \\ $$$$\Delta\:=\left(\mathrm{4}+{i}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{5}+\mathrm{6}{i}\right)\:=\mathrm{16}+\mathrm{8}{i}−\mathrm{1}−\mathrm{20}−\mathrm{24}{i}\:=−\mathrm{5}+\mathrm{16}{i}\:\:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{4}+{i}+\sqrt{−\mathrm{5}+\mathrm{16}{i}}}{\mathrm{2}}\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{4}+{i}−\sqrt{−\mathrm{5}+\mathrm{16}{i}}}{\mathrm{2}} \\ $$$${we}\:{see}\:{there}\:{no}\:{relation}\:{between}\:{the}\:\mathrm{2}\:{Questions}.. \\ $$
Commented by Tawa1 last updated on 10/Nov/18
God bless you sir. The correct question should be. z^2 − (4 + i)z + (5 + 5i) = 0. Please check sir. God bless you
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\:\mathrm{The}\:\mathrm{correct}\:\mathrm{question}\:\mathrm{should}\:\mathrm{be}. \\ $$$$\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{4}\:+\:\mathrm{i}\right)\mathrm{z}\:+\:\left(\mathrm{5}\:+\:\mathrm{5i}\right)\:=\:\mathrm{0}.\:\:\:\:\:\mathrm{Please}\:\mathrm{check}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$
Commented by maxmathsup by imad last updated on 10/Nov/18
let z^2 −(4+i)z +5+5i =0 Δ =(−(4+i))^2 −4(5+5i)=16+8i−1−20−20i=−5−12i and the roots of this equation are z_k =+^− (√(13)) e^(i((π/2) +(1/2)arctan(((12)/5)))) =+^− i (√(13)) e^((i/2)arctan(((12)/5))) .
$${let}\:{z}^{\mathrm{2}} −\left(\mathrm{4}+{i}\right){z}\:+\mathrm{5}+\mathrm{5}{i}\:=\mathrm{0} \\ $$$$\Delta\:=\left(−\left(\mathrm{4}+{i}\right)\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{5}+\mathrm{5}{i}\right)=\mathrm{16}+\mathrm{8}{i}−\mathrm{1}−\mathrm{20}−\mathrm{20}{i}=−\mathrm{5}−\mathrm{12}{i}\:{and}\:{the}\:{roots} \\ $$$${of}\:{this}\:{equation}\:{are}\:{z}_{{k}} =\overset{−} {+}\sqrt{\mathrm{13}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)\right)} \:\:=\overset{−} {+}{i}\:\sqrt{\mathrm{13}}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)} \:. \\ $$
Commented by Tawa1 last updated on 10/Nov/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Smail last updated on 10/Nov/18
−5−12i=13(((−5)/(13))−((12)/(13))i)=−13e^(itan^(−1) (((12)/5))) =13e^(i(π+tan^(−1) (((12)/5)))) (−5−12i)^(1/2) =(√(13))e^(i((π+tan^(−1) (((12)/5)))/2)+kπ) =(√(13))ie^(i((tan^(−1) (((12)/5)))/2)) or =−(√(13))ie^(i((tan^(−1) (((12)/5)))/2))
$$−\mathrm{5}−\mathrm{12}{i}=\mathrm{13}\left(\frac{−\mathrm{5}}{\mathrm{13}}−\frac{\mathrm{12}}{\mathrm{13}}{i}\right)=−\mathrm{13}{e}^{{itan}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{5}}\right)} \\ $$$$=\mathrm{13}{e}^{{i}\left(\pi+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{5}}\right)\right)} \\ $$$$\left(−\mathrm{5}−\mathrm{12}{i}\right)^{\mathrm{1}/\mathrm{2}} =\sqrt{\mathrm{13}}{e}^{{i}\frac{\pi+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{5}}\right)}{\mathrm{2}}+{k}\pi} \\ $$$$=\sqrt{\mathrm{13}}{ie}^{{i}\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{5}}\right)}{\mathrm{2}}} {or}\:=−\sqrt{\mathrm{13}}{ie}^{{i}\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{5}}\right)}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Nov/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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