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Question Number 137117 by MathZa last updated on 29/Mar/21

Answered by mathmax by abdo last updated on 30/Mar/21

is z+i(√2)=0?  z+i(√2)=((2−(√2)+(√2)i)/(1+(√2)i−i)) +(√2)i =((2−(√2)+(√2)i+(√2)i−2+(√2))/(1+(√2)i −i))  =((2(√2)i)/(1+((√2)−1)i)) =((2(√2)i(1−((√2)−1)i)/(1+((√2)−1)^2 )) =((2(√2)i+2(√2)((√2)−1))/(1+((√2)−1)^2 ))  =((2(√2)i+4−2(√2))/(4−2(√2)))≠0 ⇒z≠−(√2)i

$$\mathrm{is}\:\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\mathrm{0}? \\ $$$$\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{i}}\:+\sqrt{\mathrm{2}}\mathrm{i}\:=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}\:−\mathrm{i}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}}\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}\left(\mathrm{1}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}\right.}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}+\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}+\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\neq\mathrm{0}\:\Rightarrow\mathrm{z}\neq−\sqrt{\mathrm{2}}\mathrm{i} \\ $$

Commented by MathZa last updated on 30/Mar/21

Answered by MJS_new last updated on 30/Mar/21

((2−(√2)+(√2)i)/(1−(1−(√2))i))=(((2−(√2)+(√2)i)(1+(1−(√2))i))/((1−(1−(√2))i)(1+(1−(√2))i)))=  =((4−2(√2)+(4−2(√2))i)/(4−2(√2)))=1+i

$$\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}\right)\left(\mathrm{1}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)}{\left(\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)\left(\mathrm{1}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)}= \\ $$$$=\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}+\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{i}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{1}+\mathrm{i} \\ $$

Commented by MathZa last updated on 30/Mar/21

thank you

$${thank}\:{you}\: \\ $$

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