Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com