Proof for roots of z = a+ib (1 Viewer)

[iamgonnaflunklol]

Hi all,

Just looking for some help with the following question, cheers.

Let z = a+ib. Prove that there are always two sqaure roots of z, except for when a=b=0.

 

[大きい男]

Try forming a quadratic equation in terms of x^2 using the same process you would when finding the square root of a complex number, but using a and b as your real and imaginary parts.

 

[catietravers3]

maybe try (x+iy)^2=(a+ib) and solve that using imaginary and real parts

 

[fan96]

Here's a different style of proof with less algebraic manipulation.

You should know what the Fundamental Theorem of Algebra is, from the Polynomials topic.
To recap, the FTA essentially states that every (non-constant) complex polynomial has at least one root (in ).

So let be any complex number, and consider the polynomial

.

You can apply the FTA on .
Note that if is a root of , then so is .

 

[iamgonnaflunklol]

Try forming a quadratic equation in terms of x^2 using the same process you would when finding the square root of a complex number, but using a and b as your real and imaginary parts.

Thanks!

 

[iamgonnaflunklol]

Here's a different style of proof with less algebraic manipulation.

You should know what the Fundamental Theorem of Algebra is, from the Polynomials topic.
To recap, the FTA essentially states that every (non-constant) complex polynomial has at least one root (in ).

So let be any complex number, and consider the polynomial

.

You can apply the FTA on .
Note that if is a root of , then so is .

That makes a lot of sense, actually. And it removes the need for plenty of extra algebra too.

 

[iamgonnaflunklol]

maybe try (x+iy)^2=(a+ib) and solve that using imaginary and real parts

Thank you. Very practical.