Question Number 137117 by MathZa last updated on 29/Mar/21
Answered by mathmax by abdo last updated on 30/Mar/21
$$\mathrm{is}\:\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\mathrm{0}? \\ $$$$\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{i}}\:+\sqrt{\mathrm{2}}\mathrm{i}\:=\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}+\sqrt{\mathrm{2}}\mathrm{i}−\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{i}\:−\mathrm{i}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}}\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}\left(\mathrm{1}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{i}\right.}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}+\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}+\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\neq\mathrm{0}\:\Rightarrow\mathrm{z}\neq−\sqrt{\mathrm{2}}\mathrm{i} \\ $$
Commented by MathZa last updated on 30/Mar/21
Answered by MJS_new last updated on 30/Mar/21
$$\frac{\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\mathrm{i}\right)\left(\mathrm{1}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)}{\left(\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)\left(\mathrm{1}+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{i}\right)}= \\ $$$$=\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}+\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\mathrm{i}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{1}+\mathrm{i} \\ $$
Commented by MathZa last updated on 30/Mar/21
$${thank}\:{you}\: \\ $$