[((1)/(1-2i)+(3)/(1+i))((3+4i)/(2-4i))=(1+9i)/(4)],[(a)" True "quad " (b) False "]

(1)/(1-2i)+(3)/(1+4i)

(v) (1-i) ^ (2) (1 + i) - (3-4i) ^ (2)

((1+i)/(1-i))^(4)+((1-i)/(1+i))^(4)=(A)1(B)2(C)3(D)4

Express (1 /(2 -2i)+3/(1+i)) ((3+ 4i)/(2-4i)) in the form of a +ib

Express (1 /(2 -2i)+3/(1+i)) ((3+ 4i)/(2-4i)) in the form of a +ib

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

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

Reduce ((1)/(1-4i)-(2)/(1+i))((3-4i)/(5+i)) to the standard form.

The modulus of ((1+2i)(3-4i))/((4+3i)(2-3i)) is

If i=sqrt(-)1, then 4+5(-(1)/(2)+(i sqrt(3))/(2))^(334)+3(-(1)/(2)+(i sqrt(3))/(2))^(365) is equal to (1)1-i sqrt(3)(2)-1+i sqrt(3)(3)i sqrt(3)(4)-i sqrt(3)