For `i^2= -1, (1 - 3i)^3= ?`

Perform the following by the indicated operations. Express the result in the form x + iy,where x, y are real numbers i = sqrt(-1) : (-2-1/3i)^3 .

(1-i)(1+2i)(1-3i)=

If z = |{:(1 , 1 + 2i, - 5i),(1 - 2i,-3,5+3i),(5i,5-3i,7):}| , then (i= sqrt(-1))

Simplify: [1 / (1 - 2i) + 3 / (1 + i)] [(3 + 4i) / (2 - 4i)]

abs(1/((3 +i)^2)-1/((3-i)^2)) =

Reduce to the form A + IB ((1-i)^2-(1+i)^2)/((1-i)^3+(1+i)^3)

Express the following in the standard form a+i b :(1/(1-2i)+3/(1+i))((3+4i)/(2-4i))

Express each one of the following in the standard form a+i b :(1/(1-2i)+3/(1+i))((3+4i)/(2-4i))