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mixer [17]
3 years ago
9

Assume z = x + iy, then find a complex number z satisfying the given equation. d. 2z8 – 2z4 + 1 = 0

Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0

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°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}

Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}

Step-by-step explanation:

I start with a variable substitution:

Z^{4} = X

Then:

2X^{2}-2X+1=0

Solving the quadratic equation:

X_{1} =\frac{2+\sqrt{4-4*2*1} }{2*2} \\X_{2} =\frac{2-\sqrt{4-4*2*1} }{2*2}

X=\left \{ {{0,5+0,5i} \atop {0,5-0,5i}} \right.

Replacing for the original variable:

Z=\sqrt[4]{0,5+0,5i}

or Z=\sqrt[4]{0,5-0,5i}

Remembering that complex numbers can be written as:

Z=a+ib=|Z|e^{ic}

Using this:

Z=\left \{ {{{\frac{\sqrt{2}}{2} e^{i45°} } \atop {{\frac{\sqrt{2}}{2} e^{i-45°} }} \right.

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.

The possible angle respond to:

RAng_{12...n} =\frac{Ang +360*(i-1)}{n}

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.

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