Prove that
$e^\bar z$ = $\overline {e^z}$
where $z = a + bi $
thanks !
Complex Analysis exponential proof
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complex-analysis
complex-numbers
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0Use Euler's formula. – 2017-01-17
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0do you mean $e^{\bar z}=\overline{e^z}$? – 2017-01-17
2 Answers
1
The key point is that the conjugation is a continuous map on $\mathbb{C}$ so that it commutes with the limit and: $$\overline{\sum_{k=0}^{+\infty}\frac{z^n}{n!}}=\sum_{k=0}^{+\infty}\frac{\overline{z}^n}{n!}.$$ I also used the fact that the conjugate of a real number is itself and the conjugate of a product is the product of conjugates.
1
Using Euler's formula we have
$$ e^{\overline z} = e^{a}e^{-ib} = e^a(\cos {(-b)} + i\sin{(-b)}) = e^a(\cos {b} - i\sin{b}) $$ $$ \overline {e^z} = \overline {e^a(\cos b + i\sin b)} = e^a(\cos {b} - i\sin{b}) $$