Solving the equation (1 + x) ^ 5 = (1-z) ^ 5 about complex Z

Solving the equation (1 + x) ^ 5 = (1-z) ^ 5 about complex Z

(1+x)^5=(1-z)^5=ρ^5*e^(5iθ)
1+x=ρ*e^[i(2kπ+5θ)/5]
1+x
=ρ*e^[i(2kπ+5θ)/5]
=ρ(e^iθ)*e^(i2kpi/5)
=(1-z)e^(i2kpi/5)

Solutions of the equation 2x-3i = 3xi + 1 in the complex range

2x-3i=3xi+1
2x-3xi=1+3i
x(2-3i)=1+3i
x=(1+3i)/(2-3i)
x=(1+3i)(2+3i)/(2^2+3^2)=(2-9+9i)/13=(-7+9i)/13=(-7/13)+(9/13)i
x = (-7/13) + (9/13)i

The complex numbers of two different points a and B on the complex plane are α and β respectively. If α square + β square = 0, then what triangle is △ ABO
Hope to give me a more detailed answer, thank you

α square + β square = 0
That is, α & # 178; = - β & # 178;
Ψ α = β * I or α = β * (- I)
Using the trigonometric form of complex number, then
∠ AOB = 90 ° and | OA | = | ob|
That is to say, △ ABO is an isosceles right triangle

In the complex plane, the complex number corresponding to a is 3 + I, and the complex number corresponding to B is - 1 + 3I, then the shape of triangle AOB is
A isosceles triangle b right triangle
C isosceles right triangle D obtuse triangle

Choose C
Obviously, OA = ob = √ (1 & sup2; + 3 & sup2;) = √ 10
Vector OA = (3,1), vector ob = (- 1,3)
OA*OB=3*(-1)+1*3=0
So OA ⊥ ob