Put in polar form [(6+8i)/(4-3i)]power 2 pls show steps

Anon E. Moose has a perfect working of it in Cartesian form, converted to polar at the end. If you like his answer, you should go with it.

But here's an alternative, both for a little variety, and to help you get familiar with dealing with complex operations on their polar forms.
Convert to polar at the start, then work that through:
[cis θ
is short for
cos θ + i sin θ]

z1 = 6 + 8i = r1 cis(θ1)
r1^2 = 6^2 + 8^2 = 100; r1 = 10
θ1 = atan(8/6) = atan(4/3)
z1 = 10 cis(4/3)

z2 = 4 - 3i = r2 cis(θ2)
r2^2 = 4^2 + (-3)^2 = 25; r2 = 5
θ2 = atan(-3/4) = -atan(3/4)
z2 = 5 cis(3/4)

z1/z2 = (r1 cis(θ1))/(r2 cis(θ2))
= (r1/r2)cis(θ1 - θ2)
= 2 cis(atan(4/3) + atan(3/4))

But angles whose tangents are reciprocal of each other, are complementary:
atan(x) + atan(1/x) = π/2

So
z1/z2 = 2 cis(π/2)

And what we want is the square of this, so
(z1/z2)^2 = 2^2 cis(2*π/2) = 4 cis(π)

[(6 + 8i) / (4 - 3i)]²
Multiply and divide by the complex conjugate of the denominator inside the brackets:
[(6 + 8i)(4 + 3i) / (4 - 3i)(4 + 3i)]²
Distribute:
[(24 + 32i + 18i + 24i²) / (4² - (3i)²)]²
Simplify (i² = -1):
[(24 + 50i - 24) / (16 - (-9))]²
Combine Like Terms:
[50i / 25]²
Reduce:
[2i]²
Apply the Exponent:
4i²
Simplify (i² = -1):
-4

z = x + iy
z = -4 + 0i
r = 4
θ = π

a + bi

r = √(a^2+b^2)

θ = tan^-1(b/a)

a = r cos(θ) and b = r sin(θ)

for 6 + 8i , r = √(36+64) = 10

θ = tan^-1(8/6) = 59π/200

6 + 8i = 10 cos(59π/200) + 10i sin(59π/200)

for 4 - 3i, r = √(16+9) = 5

θ = tan^-1(-3/4) = 449π/250 (IV quadrant)

4 - 3i = 5 cos(449π/250) + 5i sin(449π/250)

=> [ (6 + 8i) /(4 - 3i) ]^2 = (6 + 8i)^2 (4 - 3i)^-2

= (10 cos(59π/200) + 10i sin(59π/200)^2 *( 5 cos(449π/250) + 5i sin(449π/250)

= (100/25) cos(59π/100) + i sin(59π/100)*( cos(449π/125) - i sin(449π/125)

= 4 cos(59π/100)cos(449π/125)+ sin(59π/100)sin(449π/125)+i{(sin(59π/100… ]

= 4 cos(59π/100 - 449π/125)+ +i{(sin(59π/100 - 449π/125}

= 4 [ cos(1501π/500) - i(sin(1501π/500) ]

roughly 4[ cos(3π) - i sin(3π) ]

= - 4 - 0

= -4

you should consider asking your teacher for this tye of advice. sorry. - piggy wiggy