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Question Number 13982 by tawa tawa last updated on 26/May/17
Find the root of the equation z^2 − 8(1 − i)z + 63 − 16i = 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{8}\left(\mathrm{1}\:−\:\mathrm{i}\right)\mathrm{z}\:+\:\mathrm{63}\:−\:\mathrm{16i}\:=\:\mathrm{0} \\ $$
Answered by ajfour last updated on 26/May/17
let z=x+iy x^2 −y^2 +2ixy−8x−8iy+8ix−8y +63−16i =0 ⇒ x^2 −y^2 −8x−8y+63=0 , and xy+4x−4y−8=0 x(y+4)−4(y+4)+8=0 (y+4)(4−x)=8 ....(i) (x−4)^2 +63=(y+4)^2 ....(ii) ⇒ (y+4)^2 =((64)/((y+4)^2 ))+63 let t=(y+4)^2 t=((64)/t)+63 t^2 −63t−64=0 t^2 −64t+t−64=0 t(t−64)+(t−64)=0 (t−64)(t+1)=0 as t+1>0 so, t=(y+4)^2 =64 y=−4±8 y=−12, 4 from (i): x=4−(8/((y+4))) =5, 3 (x,y)≡(5,−12) , (3,4) z_1 =5−12i ; z_2 =3+4i .
$${let}\:{z}={x}+{iy} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{ixy}−\mathrm{8}{x}−\mathrm{8}{iy}+\mathrm{8}{ix}−\mathrm{8}{y} \\ $$$$\:\:\:\:\:\:\:+\mathrm{63}−\mathrm{16}{i}\:=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{8}{y}+\mathrm{63}=\mathrm{0}\:\:,\:{and} \\ $$$$\:\:\:\:\:\:{xy}+\mathrm{4}{x}−\mathrm{4}{y}−\mathrm{8}=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:{x}\left({y}+\mathrm{4}\right)−\mathrm{4}\left({y}+\mathrm{4}\right)+\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left({y}+\mathrm{4}\right)\left(\mathrm{4}−{x}\right)=\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$$\:\:\:\:\:\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{63}=\left({y}+\mathrm{4}\right)^{\mathrm{2}} \:\:\:….\left({ii}\right) \\ $$$$\Rightarrow\:\:\left({y}+\mathrm{4}\right)^{\mathrm{2}} =\frac{\mathrm{64}}{\left({y}+\mathrm{4}\right)^{\mathrm{2}} }+\mathrm{63} \\ $$$${let}\:\boldsymbol{{t}}=\left({y}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{t}=\frac{\mathrm{64}}{{t}}+\mathrm{63} \\ $$$$\:\:\:{t}^{\mathrm{2}} −\mathrm{63}{t}−\mathrm{64}=\mathrm{0} \\ $$$$\:\:\:{t}^{\mathrm{2}} −\mathrm{64}{t}+{t}−\mathrm{64}=\mathrm{0} \\ $$$$\:\:\:{t}\left({t}−\mathrm{64}\right)+\left({t}−\mathrm{64}\right)=\mathrm{0} \\ $$$$\:\:\:\left({t}−\mathrm{64}\right)\left({t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${as}\:{t}+\mathrm{1}>\mathrm{0}\:\:{so}, \\ $$$$\:\:{t}=\left({y}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{64} \\ $$$$\:\:{y}=−\mathrm{4}\pm\mathrm{8} \\ $$$$\:\:\boldsymbol{{y}}=−\mathrm{12},\:\mathrm{4}\: \\ $$$${from}\:\left({i}\right): \\ $$$$\:\:{x}=\mathrm{4}−\frac{\mathrm{8}}{\left({y}+\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=\mathrm{5},\:\mathrm{3} \\ $$$$\left({x},{y}\right)\equiv\left(\mathrm{5},−\mathrm{12}\right)\:,\:\left(\mathrm{3},\mathrm{4}\right) \\ $$$$\:\:\boldsymbol{{z}}_{\mathrm{1}} =\mathrm{5}−\mathrm{12}\boldsymbol{{i}}\:; \\ $$$$\:\:\:\boldsymbol{{z}}_{\mathrm{2}} =\mathrm{3}+\mathrm{4}\boldsymbol{{i}}\:. \\ $$$$ \\ $$
Commented by ajfour last updated on 26/May/17
it is correct now !
$${it}\:{is}\:{correct}\:{now}\:! \\ $$
Commented by tawa tawa last updated on 26/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/May/17
Δ=64(1−i)^2 −4(63−16i)= 64(1−2i−1)−252+64i=−64i−252= =−4(16i+63)=−4(8^2 +2×8i−1)= =−4(8+i)^2 z=((8(1−i)±2i(8+i))/2)= { (((8−8i+16i−2)/2)),(((8−8i−16i+2)/2)) :} ⇒z= { ((3+4i)),((5−12i)) :}
$$\Delta=\mathrm{64}\left(\mathrm{1}−{i}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{63}−\mathrm{16}{i}\right)= \\ $$$$\mathrm{64}\left(\mathrm{1}−\mathrm{2}{i}−\mathrm{1}\right)−\mathrm{252}+\mathrm{64}{i}=−\mathrm{64}{i}−\mathrm{252}= \\ $$$$=−\mathrm{4}\left(\mathrm{16}{i}+\mathrm{63}\right)=−\mathrm{4}\left(\mathrm{8}^{\mathrm{2}} +\mathrm{2}×\mathrm{8}{i}−\mathrm{1}\right)= \\ $$$$=−\mathrm{4}\left(\mathrm{8}+{i}\right)^{\mathrm{2}} \\ $$$${z}=\frac{\mathrm{8}\left(\mathrm{1}−{i}\right)\pm\mathrm{2}{i}\left(\mathrm{8}+{i}\right)}{\mathrm{2}}=\begin{cases}{\frac{\mathrm{8}−\mathrm{8}{i}+\mathrm{16}{i}−\mathrm{2}}{\mathrm{2}}}\\{\frac{\mathrm{8}−\mathrm{8}{i}−\mathrm{16}{i}+\mathrm{2}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow{z}=\begin{cases}{\mathrm{3}+\mathrm{4}{i}}\\{\mathrm{5}−\mathrm{12}{i}}\end{cases} \\ $$
Commented by tawa tawa last updated on 26/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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