Answer: complex equations has n number of solutions, been n the equation degree. In this case:
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}[/tex]
[tex]Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}[/tex]
Step-by-step explanation:
I start with a variable substitution:
[tex]Z^{4} = X[/tex]
Then:
[tex]2X^{2}-2X+1=0[/tex]
Solving the quadratic equation:
[tex]X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}[/tex]
[tex]X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.[/tex]
Replacing for the original variable:
[tex]Z=\sqrt[4]{0,5+0,5i}[/tex]
or [tex]Z=\sqrt[4]{0,5-0,5i}[/tex]
Remembering that complex numbers can be written as:
[tex]Z=a+ib=|Z|e^{ic}[/tex]
Using this:
[tex]Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.[/tex]
Solving for the modulus and the angle:
[tex]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.[/tex]
The possible angle respond to:
[tex]RAng_{12...n} =\frac{Ang +360*(i-1)}{n}[/tex]
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.
