Answer: complex equations has n number of solutions, been n the equation degree. In this case:
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi11%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi101%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi191%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi281%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi78%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi168%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi258%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi348%2C75%C2%B0%7D)
Step-by-step explanation:
I start with a variable substitution:

Then:

Solving the quadratic equation:


Replacing for the original variable:
![Z=\sqrt[4]{0,5+0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5%2B0%2C5i%7D)
or ![Z=\sqrt[4]{0,5-0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5-0%2C5i%7D)
Remembering that complex numbers can be written as:

Using this:

Solving for the modulus and the angle:
![Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.](https://tex.z-dn.net/?f=Z%3D%5Cleft%20%5C%7B%20%7B%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi45%7D%7D%20%7D%20%5Catop%20%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi-45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi-45%7D%7D%20%7D%7D%20%5Cright.)
The possible angle respond to:

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"
In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º
Obtaining 8 different angles, therefore 8 different solutions.
<span>The rocket that will reach a greater distance above the ground at its maximum height is y = -4x</span>² + 80x. The answer is letter B. This is because of the coefficient presented in the x² term. The number -4 is much greater than -6. Graphing these equations into a graphing paper or excel, you would get a parabola.
Answer:
identity property of division
Step-by-step explanation:
Answer:
Therefore, the exact value of Tan 60 degrees is √3. We can also derive the values of tan 0°, 30°, 45°, 90°, 180°, 270° and 360° in the same way
Answer: x = -30
Step-by-step explanation:
y = -20 by using process of elimination substitute y as -20 in either of the equations you get -30 for x