Question-137117

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 \sqrt{2} = 0 ?

z + i \sqrt{2} = \frac{2 - \sqrt{2} + \sqrt{2} i}{1 + \sqrt{2} i - i} + \sqrt{2} i = \frac{2 - \sqrt{2} + \sqrt{2} i + \sqrt{2} i - 2 + \sqrt{2}}{1 + \sqrt{2} i - i}

= \frac{2 \sqrt{2} i}{1 + (\sqrt{2} - 1) i} = \frac{2 \sqrt{2} i (1 - (\sqrt{2} - 1) i}{1 + {(\sqrt{2} - 1)}^{2}} = \frac{2 \sqrt{2} i + 2 \sqrt{2} (\sqrt{2} - 1)}{1 + {(\sqrt{2} - 1)}^{2}}

= \frac{2 \sqrt{2} i + 4 - 2 \sqrt{2}}{4 - 2 \sqrt{2}} \neq 0 \Rightarrow z \neq - \sqrt{2} 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} i}{1 - (1 - \sqrt{2}) i} = \frac{(2 - \sqrt{2} + \sqrt{2} i) (1 + (1 - \sqrt{2}) i)}{(1 - (1 - \sqrt{2}) i) (1 + (1 - \sqrt{2}) i)} =

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

Commented by MathZa last updated on 30/Mar/21

t h a n k y o u

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