If $\sqrt {x+iy}=a+bi$ then prove that

Question

If $\sqrt {x+iy}=a+bi$ then prove that $\sqrt {x-iy}=a-ib$

My Attempt $$\sqrt {x+iy}=a+bi$$ Squaring both sides $$x+iy=a^2+2iab+i^2.b^2$$ $$x+iy=a^2-b^2+2iab$$

How do I proceed further?

Answer 1

I suppose that your question is:

If $x,y,a,b\in\mathbb R$ and if $a+bi$ is a square root of $x+yi$, then prove that $a-bi$ is a square root of $x-yi$.

If so, note that\begin{align}a+bi\text{ is a square root of }x+yi&\iff(a+bi)^2=x+yi\\&\iff a^2-b^2+2abi=x+yi\\&\iff\left\{\begin{array}{l}a^2-b^2=x\\2ab=y\end{array}\right.\\&\iff\left\{\begin{array}{l}a^2-b^2=x\\-2ab=-y\end{array}\right.\\&\iff a^2-b^2-2abi=x-yi\\&\iff(a-bi)^2=x-yi.\end{align}

Answer 2

Hint

Compare the real and imaginary parts to get $$x=a^2-b^2$$ and $$y=2ab$$ and use it for next question

Answer 3

The two square roots of $\,−1\,$ are indistinguishable, so any result that uses $\,i\,$ will also be true with $\,−i\,$ substituted for $\,i\,$ throughout. Technically, conjugation is an automorphism. You may think that this holds for square root of $\,2\,$, for example, but one root is positive and the other negative, although algebraically they are indistinguishable.

Answer 4

\begin{align} & x+iy=a^2-b^2+2iab \\ & x-iy=a^2-b^2-2iab = (a-ib)^2 \end{align}

Answer 5

$\overline{\sqrt{x+yi}}=\overline{a+bi}$

$\sqrt{x-yi}={a-bi}$