Easy proof don't know where i am going wrong

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The discussion revolves around verifying the equation (1+i)^(95) = (1-i)*2^(47). The original poster initially miscalculated the expression, arriving at (1+i)*2^(47) instead. A participant pointed out the need to clarify the values of r and theta for further assistance. Ultimately, the original poster realized their mistake stemmed from incorrectly interpreting sin(95pi/4) as 1/sqrt2 instead of its negative. The conversation concludes with an acknowledgment of the error and gratitude for the clarification.
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the question is show that (1+i)^(95) =(1-i)*2^(47)
we are told that z^(n) = r^(n) * e^(i*n*theta)

when i do the problem i get

(1+i)^(95) =(1+i)*2^(47)

can anybody verify whether i am right or wrong. thanks :biggrin:
 
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show me your r and theta b4 i can further help you... the question has no problem and I get (1-i)*2^47
 
stupid me... i know what it is now... my problem was that i had sin(95pi/4) = 1/sqrt2 and not the negative of that... thanks checking the problem

:approve:
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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