Answer:
Step-by-step explanation:
Mabye 7?
Exercise 1:
The easiest way to compute powers of complex numbers is to write them in the form
![z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))](https://tex.z-dn.net/?f=z%20%3D%20%5Crho%20e%5E%7Bi%5Ctheta%7D%20%3D%20%5Crho%28%5Ccos%28%5Ctheta%29%2Bi%5Csin%28%5Ctheta%29%29)
In this form, you have
![z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))](https://tex.z-dn.net/?f=z%5En%20%3D%20%5Crho%5En%20e%5E%7Bin%5Ctheta%7D%20%3D%20%5Crho%5En%28%5Ccos%28n%5Ctheta%29%2Bi%5Csin%28n%5Ctheta%29%29)
The magnitude of the number is given by
![z=a+bi \implies \rho = \sqrt{a^2+b^2}](https://tex.z-dn.net/?f=z%3Da%2Bbi%20%5Cimplies%20%5Crho%20%3D%20%5Csqrt%7Ba%5E2%2Bb%5E2%7D)
So, we have
![z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2](https://tex.z-dn.net/?f=z%3D-1%2B%5Csqrt%7B3%7Di%20%5Cimplies%20%5Crho%20%3D%20%5Csqrt%7B1%2B3%7D%3D2)
As for the angle, we have
![z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}](https://tex.z-dn.net/?f=z%3Da%2Bbi%20%5Cimplies%20%5Ctheta%20%3D%20%5Ctext%7Batan2%7D%5Cdfrac%7Bb%7D%7Ba%7D)
So, we have
![z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}](https://tex.z-dn.net/?f=z%3D-1%2B%5Csqrt%7B3%7Di%20%5Cimplies%20%5Ctheta%20%3D%20%5Ctext%7Batan2%7D%5Cleft%28%5Cdfrac%7B-%5Csqrt%7B3%7D%7D%7B-1%7D%5Cright%29%20%3D%20-%5Cdfrac%7B2%5Cpi%7D%7B3%7D)
Finally,
![z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64](https://tex.z-dn.net/?f=z%3D-1-%5Csqrt%7B3%7Di%20%3D%202%5Cleft%28%5Ccos%5Cleft%28%20-%5Cdfrac%7B2%5Cpi%7D%7B3%7D%5Cright%29%20%2B%20i%5Csin%5Cleft%28%20-%5Cdfrac%7B2%5Cpi%7D%7B3%7D%5Cright%29%5Cright%29%20%5Cimplies%20z%5E6%20%3D%202%5E6%5Cleft%28%5Ccos%5Cleft%28%20-4%5Cpi%5Cright%29%20%2B%20i%5Csin%5Cleft%28-4%5Cpi%5Cright%29%5Cright%29%20%3D%2064)
Exercise 2:
You simply have to compute the trigonometric function:
![\cos\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2},\quad \sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%20%3D%20%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2C%5Cquad%20%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%20%3D%20-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
So, we have
![z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right) = 1-i](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%7B2%7D%5Cleft%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20-%20i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cright%29%20%3D%201-i)
I am attaching a document showing the steps.
Well x=1 always unless it tells you other wise so 1+0.5 would equal 1.5 and so after that you keep adding so 1.5 + 0.5 equal 2 so you keep taking the previous answer and adding 0.5 to it
The transformation of a function may involve any change. The correct option is D.
<h3>How does the transformation of a function happen?</h3>
The transformation of a function may involve any change.
Usually, these can be shifted horizontally (by transforming inputs) or vertically (by transforming output), stretched (multiplying outputs or inputs), etc.
If the original function is y = f(x), assuming the horizontal axis is the input axis and the vertical is for outputs, then:
Horizontal shift (also called phase shift):
- Left shift by c units, y=f(x+c) (same output, but c units earlier)
- Right shift by c units, y=f(x-c)(same output, but c units late)
Vertical shift
- Up by d units: y = f(x) + d
- Down by d units: y = f(x) - d
Stretching:
- Vertical stretch by a factor k: y = k \times f(x)
- Horizontal stretch by a factor k: y = f(\dfrac{x}{k})
Given the function f(x)=2ˣ, while the h(x)=-3(2ˣ), therefore, the function f(x) is a reflection and a translation of a function. Hence, the correct option is D.
Learn more about Transforming functions:
brainly.com/question/17006186
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