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Question Number 115325 by zakirullah last updated on 25/Sep/20
Commented by mohammad17 last updated on 25/Sep/20
Q5/i) z^2 +z+3=0⇒z=((−b±(√(b^2 −4ac)))/(2a))⇒z=((−1±i(√(11)))/2) ∴z=−(1/2)−((√(11))/2)i , z=−(1/2)+((√(11))/2)i ii)z^2 −1=z⇒z^2 −z−1=0⇒z=((1±(√5))/2)⇒z=(1/2)−((√5)/2),z=(1/2)+((√5)/2) iii)z^2 −2z+i=0⇒z=((2±2(√(1−i)))/2)⇒z=1−(√(1−i)),z=1+(√(1−i)) iv)z^2 +4=0⇒(z−2i)(z+2i)=0⇒z=2i,z=−2i
$$\left.{Q}\mathrm{5}/{i}\right) \\ $$$$ \\ $$$${z}^{\mathrm{2}} +{z}+\mathrm{3}=\mathrm{0}\Rightarrow{z}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\Rightarrow{z}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$$\therefore{z}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}{i}\:\:,\:{z}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}{i} \\ $$$$ \\ $$$$\left.{ii}\right){z}^{\mathrm{2}} −\mathrm{1}={z}\Rightarrow{z}^{\mathrm{2}} −{z}−\mathrm{1}=\mathrm{0}\Rightarrow{z}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}},{z}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$$\left.{iii}\right){z}^{\mathrm{2}} −\mathrm{2}{z}+{i}=\mathrm{0}\Rightarrow{z}=\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{1}−{i}}}{\mathrm{2}}\Rightarrow{z}=\mathrm{1}−\sqrt{\mathrm{1}−{i}},{z}=\mathrm{1}+\sqrt{\mathrm{1}−{i}} \\ $$$$ \\ $$$$\left.{iv}\right){z}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\Rightarrow\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)=\mathrm{0}\Rightarrow{z}=\mathrm{2}{i},{z}=−\mathrm{2}{i} \\ $$
Commented by zakirullah last updated on 25/Sep/20
solve Q5,6
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{Q}}\mathrm{5},\mathrm{6} \\ $$
Commented by bobhans last updated on 25/Sep/20
your image not good
$${your}\:{image}\:{not}\:{good} \\ $$
Commented by zakirullah last updated on 25/Sep/20
sir click the image on the screen then you can see?
$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{click}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{image}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{screen}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{see}}? \\ $$$$ \\ $$
Commented by mohammad17 last updated on 25/Sep/20
Q6) i)z^4 +z^2 +1=0 put:z^2 =w⇒w^2 +w+1=0⇒w=((−1±i(√3))/2)⇒w=−(1/2)−((i(√3))/2)⇒z=−(1/2)+i((√3)/2) ,z=(1/2)−((√3)/2)i w=−(1/2)+((√3)/2)i⇒z=(1/2)−((√3)/2)i , z=−(1/2)−((√3)/2)i ii)(z−1)^3 =−1 put:m=z−1⇒m^3 +1=0⇒(m+1)(m^2 −m+1)=0 ⇒m+1=0⇒z−1+1=0⇒z=0 (m^2 −m+1)=0⇒m=((1±(√3))/2)i m=(1/2)−((√3)/2)i⇒z−1=(1/2)−((√3)/2)i⇒z=(3/2)−((√3)/2)i m=(1/2)+((√3)/2)i⇒z−1=(1/2)+((√3)/2)i⇒z=(3/2)+((√3)/2) i
$$\left.{Q}\mathrm{6}\right) \\ $$$$\left.{i}\right){z}^{\mathrm{4}} +{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${put}:{z}^{\mathrm{2}} ={w}\Rightarrow{w}^{\mathrm{2}} +{w}+\mathrm{1}=\mathrm{0}\Rightarrow{w}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{w}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{z}=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:,{z}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$$ \\ $$$${w}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:,\:{z}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$$ \\ $$$$\left.{ii}\right)\left({z}−\mathrm{1}\right)^{\mathrm{3}} =−\mathrm{1} \\ $$$${put}:{m}={z}−\mathrm{1}\Rightarrow{m}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\Rightarrow\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{m}+\mathrm{1}=\mathrm{0}\Rightarrow{z}−\mathrm{1}+\mathrm{1}=\mathrm{0}\Rightarrow{z}=\mathrm{0} \\ $$$$ \\ $$$$\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)=\mathrm{0}\Rightarrow{m}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${m}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\Rightarrow{z}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\Rightarrow{z}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$$ \\ $$$${m}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\Rightarrow{z}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\Rightarrow{z}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{i} \\ $$

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