let-z-a-ib-find-f-z-z-x-2-dx-

Question Number 36915 by maxmathsup by imad last updated on 07/Jun/18

l e t z = a + i b f i n d f (z) = \int_{- \infty}^{+ \infty} z^{- x^{2}} d x

Commented by math khazana by abdo last updated on 02/Aug/18

w e h a v e ∣ z ∣= \sqrt{a^{2} + b^{2}} \Rightarrow z = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} + \frac{b}{\sqrt{a^{2} + b^{2}}}) \Rightarrow

= r e^{i θ} \Rightarrow r = \sqrt{a^{2} + b^{2}} a n d t a n θ = \frac{b}{a} (w e t a k e a \neq o) \Rightarrow

θ = a r c t a n (\frac{b}{a}) \Rightarrow z = \sqrt{a^{2} + b^{2}} e^{i a r c t a n (\frac{b}{a})} \Rightarrow

f (z) = \int_{- \infty}^{+ \infty} {({r e}^{i θ})}^{- x^{2}} d x

= \int_{- \infty}^{+ \infty} {(e^{l n (r) + i θ})}^{- x^{2}} d x

\int_{- \infty}^{+ \infty} e^{- (l n (r) + i θ)) x^{2}} d x c h a n g e m e n t

\sqrt{l n (r) + i θ} x = u g i v e

f (z) = \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{l n (r) + i θ}}

= \frac{π}{\sqrt{\frac{1}{2} l n (a^{2} + b^{2}) + i a r c t a n (\frac{b}{a})}} w i h z = a + i b

a n d a \neq 0