How do I solve these 2 imaginary number problems?

Question

I can't seem to get them right!

Thanks in advanced. :)

Evaluate the expression and write the result in the form a + bi.
(√16 - √-14)(√32 - √-28)

&
Evaluate the expression and write the result in the form a + bi.
√-4√-16/√-400

Answer 1

2*sqrt(2)-(16*I)*sqrt(7)

Evaluate the expression and write the result in the form a + bi.
√-4√-16/√-400
2I*4I/20I=8I^2/20I=
8I/20=0+(2/5)I ANSWER

Answer 2

you may build 2 equations for the two unknowns x and y, then you somewhat can resolve it, i think of you have gotten suggested that. unquestionably the fee sign in complicated variables isn't a similar element it incredibly is in actual numbers (how could it, appropriate? How can an imaginary huge style be the two helpful or detrimental, it incredibly is complicated. What it capability is a significance. people use different phrases for it, a significance, an "absolute fee," a norm, a modulus, a length, perchance greater. besides, you compute a significance in complicated variables as a Pythagorean sum: |x + iy| = sqrt{ x^2 + y^2} So, |x + iy| = (y - ix)^2 sqrt{x^2 + y^2} = (y - ix)^2 = y^2 - 2 i xy - x^2 sqrt{x^2 + y^2} = y^2 - x^2 - i (2xy) Equating actual and imaginary aspects actual area: sqrt{x^2 + y^2} = y^2 - x^2 and Imaginary area: 0 = - 2xy The imaginary area calls for that the two x = 0 or y = 0, locate the two units of x and y for the two one equaling 0. ------ if x = 0 the actual area equation tells us sqrt{x^2 + y^2} = y^2 - x^2 sqrt{y^2} = y^2 y = y^2 0 = y^2 - y 0 = y(y - one million), so y = 0 and y = one million are suggestions so one attainable answer set is, x = 0, y = 0 or one million.........[Ans.] ------- if y = 0 then the actual area equation tells us sqrt{x^2 + y^2} = y^2 - x^2 sqrt{x^2} = -x^2 x = -x^2 x^2 + x = 0 x(x + one million) = 0, so x = 0 and x = -one million are suggestions for this reason, the relax answer set is: y = 0, x = 0 or -one million..........[Ans.]

Answer 3

(√16 - √-14)(√32 - √-28)
=(4-√14i)(4√2-2√7i)
=16√2-8√7i-4√28i-2√7√14
=16√2-2√7√7√2 - (8√7+8√7)i
=16√2-14√2 - 16√7i
2√2 - 16√7i

√-4√-16/√-400
=2i x 4i / 20i
= 2i/5
= 0 +2/5 i