Question

Represent the following complex no. in the standard form x+iy÷

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Answer 1

$as \: we \: know \: that \\ re {}^{ \theta i} = r( \cos( \theta) + i \sin( \theta) ) \\ question \\ \sqrt{2} (cos( \frac{ - \pi}{4} ) + i \sin( \frac{ - \pi}{4} ) ) \\ so \\ r = \sqrt{2} \\ \theta = \frac{ - \pi}{4} \\ then \\ \sqrt{2} (cos( \frac{ - \pi}{4} ) + i \sin( \frac{ - \pi}{4} ) ) \\ = \sqrt{2} \times e {}^{ \frac{ - \pi}{ 4 } i} \\ = \sqrt{2} e {}^{ \frac{ - \pi i}{4} } \: is \: your \: answer \\ if \: you \: want \: to \: find \\ the \: value \: \\ then \\ \sqrt{2} \times {({e}^{\pi i}) }^{ \frac{ - 1}{4} } \\ where \: {e}^{\pi i} = - 1 = {i}^{2} \\ \sqrt{2} \times {i}^{2 \times \frac{ - 1}{4} } \\ = \sqrt{2} \times {i}^{ \frac{ - 1}{2} } \\ = \sqrt{2} \times \frac{1}{ {}^{ {i}^{ \frac{1}{2} } } } \\ = \sqrt{2} \times \frac{1}{ {i}^{ \frac{1}{2} } } \times \frac{ {i}^{ \frac{1}{2} } }{ {i}^{\frac{1}{2} }} \\ = \sqrt{2 } \times \frac{ \sqrt{i} }{i}$