Finding dt/dθ with t2-1 = 2√(z2-1)exp(iθ)(z+√(z2-1)cosθ)

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The discussion centers on finding the derivative dt/dθ from the equation t² - 1 = 2√(z² - 1)exp(iθ)(z + √(z² - 1)cosθ) and the definition t = z + √(z² - 1)exp(iθ). The user seeks a simpler method for differentiation due to the complexity introduced by square roots. A participant clarifies that the two equations define s as a function of θ and are not equivalent, indicating a misunderstanding in the relationship between the variables.

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How do I most easily find dt/dθ, when they are related by the following equation:

t2-1 = 2√(z2-1)exp(iθ)(z+√(z2-1)cosθ)

and t = z + √(z2-1)exp(iθ)

I could find t and then differentiate, but with all the square roots it's going to be a mess. Also my teacher seemed to think it was easy, so therefore I ask what else can be done in a situation like this.

Thanks
 
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I'm not clear what you mean here. You have two different equations defining s as a function of [itex]\theta[/itex] (I presume z is to be treated as a constant.). And they are NOT equivalent.
 

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