(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)
… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)
… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))
… = (2 - 2√3 <em>i</em> ) / 4
… = 1/2 - √3/2 <em>i</em>
… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))
… = exp(<em>i</em> arctan(-√3))
… = exp(-<em>i</em> arctan(√3))
… = exp(-<em>iπ</em>/3)
By DeMoivre's theorem,
[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1
I think A, but im not 100% sure
Answer: Yes, No
Step-by-step explanation:
1st table is linear because the slope is always constant.
Second is not linear, but rather exponential.
The derivative of inverse trigonometric identities follow<span> a set of rules. </span>
<span>For </span>arc cos<span> x, the </span>dy/dx is equal to -1/ square root of (1 - x2).
<span>In this case, we still have 1/2 x inside the angle quantity so we multiply the </span>dy/dx above with 1/2. Hence, y' = is -1<span>/ 2 square root of (1 - x2); x = 1, y' is infinity. this means the line is horizontal. The answer hence is y = 1.</span>