-1

$Z = x+ iy$ Find real and imaginary part of $\sqrt{x+iy}$

I don't know how to even begin with this , I am really confused

Varun Vejalla
  • 8,873
user268439
  • 13
  • 3
  • 5
    Does this answer your question? How do I get the square root of a complex number? – Thulashitharan D Jul 10 '20 at 05:53
  • One algebraic idea is to write $(a+ib)^2=a^2-b^2+2abi=x+iy$ with $a,b\in\Bbb R$. Then you have $\begin{cases}a^2-b^2=x\ ab=\frac y2\end{cases}$. This is equivalent to the system $\begin{cases}a^2-b^2=x\ a^2(-b^2)=-\frac { y^2}4\ \operatorname{sgn}(ab)=\operatorname{sgn}y\end{cases}$. The first two equations may be solved in $a^2$ and $-b^2$ as a sum-and-product problem. The third one determines which two out of the four solutions $(a,b)$ to the system $\begin{cases}a^2-b^2=x\ a^2(-b^2)=-\frac{ y^2}4\ \end{cases}$ you may keep. –  Jul 10 '20 at 06:02

2 Answers2

1

Any complex number has two forms - either the one you've used which is a sort of cartesian form in terms of real and imaginary parts. Another is the polar form, which relies on the modulus and angle the complex number makes with the real axis

$$Z = x+iy = |Z|e^{i\theta}$$

Now, if you will see

$$Z^{1/2} = |Z|^{1/2}e^{i\frac{\theta}{2}}$$

Here

$$|Z| = \sqrt{x^2 + y^2}, \\\theta = \arctan(\frac{y}{x})$$

Hence, to re-obtain in cartesian form

$$|Z|^{1/2} = (x^2+y^2)^{\frac{1}{4}}\left(\cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right)\right)$$

EDIT

To eliminate the trigonometric functions, you could use

$$\tan \theta = \frac{y}{x} \implies \begin{cases}\sin\theta = \frac{y}{\sqrt{x^2+y^2}} \\ \cos\theta = \frac{x}{\sqrt{x^2+y^2}}\end{cases}$$

Then use the half angle formulae to find the required terms

Dhanvi Sreenivasan
  • 5,637
1

Let$$\sqrt{x+iy}=a+ib$$ Squaring on both sides $$x+iy=a^2-b^2+2iab$$Equating real and imaginary parts$$x=a^2-b^2 and\ y=2ab$$We know that $$(a^2+b^2)^2=(a^2-b^2)^2+4ab$$Notice that RHS of this equation is known and we can find the value of $(a^2+b^2)$ with which we can easily find the value of a and b by solving the system of two eqations ($a^2-b^2=x $ and $a^2+b^2=\sqrt{x^2+2y}$)
Which hence gives our answer

Thulashitharan D
  • 386