Question

[(3-i)^1/2]/1+i=x+yi... i for imaginery number, help me pls, my brain is bursting...

Answer 1

The question is...?

Answer 2

just solve this equation, find the value of x and y as they are real number

Answer 3

Take the square of both sides, the left hand side will be:
\[{3-i \over (1+i)^2}={3-i \over 2i}=-{1 \over 2}-{3 \over 2}i\]
The right hand side will be \((x+yi)^2=x^2+ x y i-y^2\). So, w have the two equations:
\(x^2-y^2=-\frac{1}{2}\) and \(xy=-\frac{3}{2}\). Solve them and get your values.

Answer 4

Don't forget to check each value you get and then reject all values that don't satisfy the original equation.

Answer 5

It should be \(2xy=-3/2\)