Complex numbers question

[umricky]

If z=1+2i is a root of the equation z^2 = az+ b find the values of a and b

I have no idea how to answer this, but I'm sure it's pretty simple.

Thanks in advance!

[Guest]

We will first find the complex conjugate of z.

Given that z = 1 + 2i, the complex conjugate of z, denoted as z*, is 1 - 2i.

Now, we'll use the property that if z is a root, its conjugate z* is also a root. Therefore, we can write the equation as:

[tex](1+2i)^2 = a(1+2i) + b[/tex] and [tex](1-2i)^2 = a(1-2i) + b[/tex]

Solving these two simultaneous equations:

Hence, [tex]a = -2 + 2i[/tex]
[tex]b = 3 + 8i[/tex]

[Guest]

You didn't specify if a and b are real or complex so I'll assume they are complex.

Let z0 be the other root. There is an infinite number of solutions for a and b :

a = 1+2i +z0, b = (1+2i)*z0, with z0 being any complex number.

If for some reason you want a and b to be real, z0 has to be equal to 1-2i and a,b=2,5 becomes the only solution.

(If a quadratic equation has real coefficients, the conjugate of any solution is also a solution, as you can see by conjugating the whole equation.) I put this in parentheses as you don't need it to solve the problem.

[Guest]

We will first find the complex conjugate of z.

Given that z = 1 + 2i, the complex conjugate of z, denoted as z*, is 1 - 2i.

Now, we'll use the property that if z is a root, its conjugate z* is also a root. Therefore, we can write the equation as:

[tex](1+2i)^2 = a(1+2i) + b[/tex] and [tex](1-2i)^2 = a(1-2i) + b[/tex]

Solving these two simultaneous equations:

Hence, [tex]a = -2 + 2i[/tex]
[tex]b = 3 + 8i[/tex]

Goodness me... If a and b aren't real numbers, z and z* cannot be both roots. Check my much simpler (and correct) solution below or above.

[Guest]

Let's solve the equation \(z^2 = az + b\) using the given root \(z = 1 + 2i\).

Substitute \(z = 1 + 2i\) into the equation:

\[(1 + 2i)^2 = a(1 + 2i) + b\]

Simplify the left side:

\[(1 + 2i)^2 = (1 + 2i)(1 + 2i) = 1 + 4i + 4i^2\]

Remember that \(i^2 = -1\), so:

\[1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i\]

Now, substitute this result back into the equation:

\[-3 + 4i = a(1 + 2i) + b\]

Distribute on the right side:

\[-3 + 4i = a + 2ai + b\]

Now, equate the real and imaginary parts on both sides:

Real part: \(-3 = a + b\)

Imaginary part: \(4 = 2a\)

Solve the system of equations to find \(a\) and \(b\):

From the imaginary part, \(2a = 4\), so \(a = 2\).

Substitute \(a = 2\) into the real part:

\[-3 = 2 + b\]

Solve for \(b\):

\[b = -5\]

So, the values of \(a\) and \(b\) are \(a = 2\) and \(b = -5\).

[JarredFunk]

Thanks for answering in brief, you made my day.

[Nessell]

thank you for your answers

[Guest]

If z=1+2i is a root of the equation z^2 = az+ b find the values of a and b

I have no idea how to answer this, but I'm sure it's pretty simple.

Thanks in advance!

a=2
b= -5.