what is the polar form of sqr -1^sqr -1?

Question

logarithm of complex numbers,polar form, trigonometric form and exponential form

Answer 1

Let (a+ib)²=i then a+ib=±√i, where a and b are both real.

a²+2abi-b²=i, so 2ab=1 and a²-b²=0 so a=±b.

(1) a=b: 2ab=1=2a²=1, b=a=±√2/2;

(2) a=-b: 2ab=1, 2a²=-1, which has no real solution, so we reject this, because a has to be real.

Therefore a=b=(i) √2/2 or (ii) -√2/2. 

(i) a²+2abi-b²=½+i-½=i or (ii) ½+i-½=i.

(√2/2)(1+i)=√i.

(√i)^√i=i^½√i=i^{(√2/4)(1+i)}.

re^iθ≡rcosθ+irsinθ, and we can write x=rcosθ and y=isinθ as Cartesian representation of the complex number, while r and θ represent the complex number in polar coordinates.

e^niθ≡cos(nθ)+isin(nθ), so ln(e^niθ)=niθ=ln(cos(nθ)+isin(nθ)); when n=1 and θ=½π, ½iπ=ln(i),

Cartesian (0,½π), polar (½π,½π).

ln(i^{(√2/4)(1+i)})=(√2/4)(1+i)ln(i)=(√2/4)(1+i)(½iπ)=⅛π√2(-1+i).

i^{(√2/4)(1+i)}=e^{⅛π√2(-1+i)}=e^-⅛π√2e^⅛iπ√2=e^-⅛π√2(cos(⅛π√2)+isin(⅛π√2),

polar coordinates (e^-⅛π√2,⅛π√2), representing (√i)^√i.

(Cartesian: (e^-⅛π√2cos(⅛π√2),e^-⅛π√2​sin(⅛π√2))).

√i is represented by (√2/2,√2/2) in Cartesian coordinates and by (1,π/4) in polar coordinates.

i is represented by (0,1) in Cartesian coordinates and by (1,π/2) in polar coordinates.

1+i⇒(1,1) Cartesian⇒rcosθ=1 and rsinθ=1⇒θ=π/4 and r=√2, (√2,π/4) polar.

[ln(i)=½iπ, ln(iⁱ)=iln(i)=-½π, e^iln(i)=e^{ln(i^i)}=e^-½π (a real number):

Cartesian and polar (e^-½π,0).]