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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

$$\mathrm{is}\mathrm{z}+\mathrm{i}\sqrt{2}=0?$$$$\mathrm{z}+\mathrm{i}\sqrt{2}=\frac{2-\sqrt{2}+\sqrt{2}\mathrm{i}}{1+\sqrt{2}\mathrm{i}-\mathrm{i}}+\sqrt{2}\mathrm{i}=\frac{2-\sqrt{2}+\sqrt{2}\mathrm{i}+\sqrt{2}\mathrm{i}-2+\sqrt{2}}{1+\sqrt{2}\mathrm{i}-\mathrm{i}}$$$$=\frac{2\sqrt{2}\mathrm{i}}{1+(\sqrt{2}-1)\mathrm{i}}=\frac{2\sqrt{2}\mathrm{i}(1-(\sqrt{2}-1)\mathrm{i}}{1+{(\sqrt{2}-1)}^2}=\frac{2\sqrt{2}\mathrm{i}+2\sqrt{2}(\sqrt{2}-1)}{1+{(\sqrt{2}-1)}^2}$$$$=\frac{2\sqrt{2}\mathrm{i}+4-2\sqrt{2}}{4-2\sqrt{2}}\ne 0\Rightarrow \mathrm{z}\ne -\sqrt{2}\mathrm{i}$$

Commented by MathZa last updated on 30/Mar/21

Answered by MJS_new last updated on 30/Mar/21

$$\frac{2-\sqrt{2}+\sqrt{2}\mathrm{i}}{1-(1-\sqrt{2})\mathrm{i}}=\frac{(2-\sqrt{2}+\sqrt{2}\mathrm{i})(1+(1-\sqrt{2})\mathrm{i})}{(1-(1-\sqrt{2})\mathrm{i})(1+(1-\sqrt{2})\mathrm{i})}=$$$$=\frac{4-2\sqrt{2}+(4-2\sqrt{2})\mathrm{i}}{4-2\sqrt{2}}=1+\mathrm{i}$$

Commented by MathZa last updated on 30/Mar/21

$$thankyou$$

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