Let z x i middot y where y 0 Choose a positive real numbe

Let z = x + i middot y, where y > 0. Choose a positive real number t so that t^2 = x^2 + y^2 and pick positive numbers u and upsilon satisfying u^2 = (t + x)/2, upsilon^2 = (t - x)/2. Show that x + i middot y = (u + i middot upsilon)^2. Can you do a similar procedure of finding a root if y is negative? Let a notequalto 0, b, and c be complex constants. Show that the quadratic equation a middot z^2 + b moddot z + c = 0 has one or two roots.

Solution

To show

x+iy =(u+i.v)²

(u+iv)² =u²-v²+2uv i

equating the real and imaginary parts

x should be equal to u²-v² and y should be equal to 2u.v ... i

given t²=x²+y² and u²=(t+x)/2 and v²=(t-x)/2

taking u²-v² from the equations given above we get

u²-v²=(t+x)/2_(t-x)/2 =2x/2=x

now take 4u²v² =4*(t+x)/2*(t-x)/2 =4*(t²-x²)/4 =1*(y²)

thus 4u²v²=y²

or, (2uv)²=(y)²

=> 2uv =y

Thus we have x =(u²-v²) and y =2u.v (proved)

Also if y is negetive then any one of u or v should also be negetive (y is a product of u and v)

But the given equations are expressed in squared terms of u and v and hence the negetivity of u or v doesnt impact the outcome of nature of other parameters related to it since they are related to squares of u and v which will always be positive regardless of sign of u and v.

Thus we can say that even if y is negetive, we can find the square roots.

We can understand the problem practically by taking x=3, y=4 and hence t=5.

u will be 2 and v will be 1 .

2). The roots of any quadratic equation is: z=b±(b²4ac)^1/2 /2a


And there are two cases to consider:

case 1: When b²4ac0 . . . We have two real roots.

The roots are: z=(b±(b²4ac)/2a)+0i


Case 2: b²4ac<0 . . . We have two complex roots.

The roots are: z=b/2a±i(4acb²)/2a