(x+3i)²=2yi

Question

Complex equation
Answer plz

Answer 1

( x + 3 i )² = x² + 2 ∙ x ∙ 3 i + ( 3 i )² =

x² + 6 x i + ( 3)² ∙ ( i )² =

x² + 6 x i + 9 ∙ ( - 1 ) =

x² + 6 x i - 9

( x + 3 i )² = 2 y i

x² + 6 x i - 9 = 2 y i

For the real parts:

x² - 9 = 0

x² = ±√ 9

x = ± 3

For the imaginary parts:

6 x i = 2 y i Divide both sidesd by 2 i

3 x = y

y = 3 x

Now:

x = - 3

y = 3 x = 3 ∙ ( - 3 ) = - 9

x = 3 x

y = 3 ∙ 3 = 9

Solutions:

x = - 3 , y = - 9

and

x = 3 , y = 9

Answer 2

Remark:

For the real parts:

x² - 9 = 0

becouse:

x² + 6 x i - 9 = 2 y i

you can write as:

x² + 6 x i - 9 = 0 + 2 y i

So the real parts is:

x² - 9 = 0

Answer 3

(x+3i)²=2yi

Answer 4

If a,b are the roots of pt²+qt+q=0prove that

Answer 5

We need to prove that a+b = -q/p.

By using the quadratic formula, we know that the roots of the equation pt² + qt + r = 0 are given by:

t = (-q ± √(q² - 4pr)) / 2p

Substituting t = a and t = b, we get:

a = (-q ± √(q² - 4pr)) / 2p

b = (-q ∓ √(q² - 4pr)) / 2p

Note that a and b have opposite signs in front of the square root because they are different roots.

Adding these two equations, we get:

a + b = (-2q) / 2p = -q/p

Therefore, we have proved that a+b = -q/p.