Find-i-i-

Question Number 88678 by Cheyboy last updated on 12/Apr/20

F i n d \sqrt{i} + \sqrt{- i}

Commented by mr W last updated on 12/Apr/20

b o t h d e f i n i t i o n s a r e u s e d (i n d i f f e r e n t

c o u n t r i e s) :

- π < A r g (z) ⩽ π o r

0 ⩽ A r g (z) < 2 π

Commented by mr W last updated on 12/Apr/20

\sqrt{i} + \sqrt{- i}

= {(e^{\frac{π i}{2}})}^{\frac{1}{2}} + {(e^{\frac{3 π i}{2}})}^{\frac{1}{2}}

= e^{\frac{π i}{4}} + e^{\frac{3 π i}{4}}

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

= \sqrt{2} i

Commented by Cheyboy last updated on 12/Apr/20

T h a n k y o u r s i r

Commented by Tony Lin last updated on 12/Apr/20

i = c o s \frac{π}{2} + i s i n \frac{π}{2}

\sqrt{i} = c o s \frac{π}{4} + i s i n \frac{π}{4}

- i = c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})

\sqrt{- i} = c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})

\sqrt{i} + \sqrt{- i} = c o s \frac{π}{4} + c o s (- \frac{π}{4}) + i [s i n \frac{π}{4} + s i n (- \frac{π}{4})]

= \sqrt{2}

Commented by ajfour last updated on 12/Apr/20

- π < A r g (z) ⩽ π

i s t h i s t r u e, s i r ?

Commented by mr W last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

t h a n k s S i r .

Commented by mathmax by abdo last updated on 12/Apr/20

\sqrt{i} + \sqrt{- i} = {(e^{\frac{i π}{2}})}^{\frac{1}{2}} + {(e^{- \frac{i π}{2}})}^{\frac{1}{2}} = e^{\frac{i π}{4}} + e^{- \frac{i π}{4}} = 2 c o s (\frac{π}{4}) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}