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Nataliya [291]
3 years ago
9

ss="latex-formula">

z = ____ + ____ i
Mathematics
1 answer:
abruzzese [7]3 years ago
3 0

If <em>z</em> ⁷ = 128<em>i</em>, then there are 7 complex numbers <em>z</em> that satisfy this equation.

z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}

\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}

(where <em>n</em> = 0, 1, 2, …, 6)

\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}

\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)

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Given quadrilateral EFGH at e(-4,7) f(8,4) G(t,-5) and h(-7,-1) using coordinate geometry prove EFGH is a rectangle
Tcecarenko [31]

The above question was not properly written.

Complete Question

Given quadrilateral EFGH with vertices at E(-4,8), F(8,4), G(5,-5) and H(-7,-1), prove using coordinate geometry that EFGH is a rectangle.

Answer:

Quadrilateral EFGH is a rectangle.l because:

EF = GH and FG = EH

Step-by-step explanation:

The formula for coordinate geometry is given as :

√(x2 - x1)² + (y2 - y1)² when we have coordinates: (x1, y1) and (x2 , y2)

For the quadrilateral EFGH with given coordinates above to be a rectangle,

EF = GH

FG = EH

Hence:

For side EF

E(-4,8), F(8,4)

= √(8 - (-4))² + (4 - 8)²

= √12² + -4²

= √144 + 16

= √160 units

For side FG

F(8,4), G(5,-5)

=√(5 - 8)² + (-5 - 4)²

= √-3² + -9²

= √9 + 81

= √90 units

For Side GH

G(5,-5) , H(-7,-1)

= √(-7 - 5)² + (-1 - (-5))²

= √-12² + 4²

= √144 + 16

= √160 units

For side EH

E(-4,8), H(-7,-1)

= √(-7 -(-4))² +(-1 - 8)²

= √-3² + -9²

= √9 + 81

= √90 units

From the above calculation, we can see that truly,

EF = GH

FG = EH

Therefore, quadrilateral EFGH is a rectangle.

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