Find the polar form of a complex number

[trv]

Homework Statement

Hi, I'm working through an example 6.3(c) from Mathematics for Physicists, by Martin and Shaw but am unable to follow some of the steps.

Question: Find the polar forms of the complex number $z = 3-\sqrt{i}$

Relevant Equations

N/A

The working out suggests first equating $\sqrt{i} = x + iy$ and suggests that squaring and equating real and imaginary parts of both sides results in $\sqrt{i} = \pm (1+i)/ \sqrt{2}$

Squaring both sides results in:
$$ i = (x + iy)^2 $$
$$ i = x^2 + 2ixy -y^2 $$

equating real parts gives
$$ x^2 - y^2 = 0 $$
$$ (x+y)(x-y) = 0 $$
$$ x = \pm y $$

equating imaginary parts gives:
$$ i = 2ixy $$
$$ 2xy = 1 $$

I'm not really sure how to proceed from here.

 

[fresh_42]

Just go ahead. If $x=\pm y$ and $2xy=1$ then $2xy=\pm 2y^2=1$ and $y=\pm 1/\sqrt{2}$ since $y$ is real. Note that squaring created two solutions, which had been only one solution before, so only one sign of $x=\pm y$ is correct.

 

[FactChecker]

Note that squaring created two solutions, which had been only one solution before, so only one sign of $x=\pm y$ is correct.

In a physics application, I would be cautious about ruling out either solution unless the physics that led to the OP forced one branch of the square root to be invalid. Either one, or both might apply. Continue to work with both until the final answer is obtained and then see which one(s) work in the physical problem.

 

[Mark44]

An easier way to approach the problem you're working, which is to find $\sqrt i$, is to think in terms of the polar form of i, which is $i = 1e^{i\pi/2}$. Here the magnitude is 1 and the argument (or angle) is $\frac \pi 2$. One square root of i will be such that its magnitude is the square root of i's magnitude, or just 1 again, and its argument will be half of the argument of i, or $\frac \pi 4$. So in polar form $\sqrt i = 1e^{i\pi/4} = 1(\cos(\pi/4) + i\sin(\pi/4)) =\frac{\sqrt 2} 2 + i\frac{\sqrt 2} 2$.
The other square root will have an argument of $\frac{5\pi}4$.

Now that you have a value for $\sqrt i$, you can finish the problem you started with; namely, finding the polar forms of $z = 3 - \sqrt i$.

 

[fresh_42]

In a physics application, I would be cautious about ruling out either solution unless the physics that led to the OP forced one branch of the square root to be invalid. Either one, or both might apply. Continue to work with both until the final answer is obtained and then see which one(s) work in the physical problem.

Yes, my basic remark was that squaring isn't an equivalence operation, i.e., it cannot be uniquely reversed. One has to keep this in mind during calculations. But your advice is a nice one: operate with all solutions until the end and then decide which ones make sense and which do not. That can help in many situations. Students often (I'd say in 99% of all cases) forget to verify if a solution they have found, e.g., by squaring steps, is actually one. Calculations are designed as steps of necessity, and the final step of satisfiability is omitted. A standard mistake in my mind.

 

[trv]

Hi all, thanks a lot for the explanations. I was able to follow the rest of the explanation in the text-book once I was able to make sense of that step.