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I sometimes find myself overcomplicating my life... overthinking simple concepts.

Here I don't use what's given, i.e., $$(a − ib)^2 = 4i$$

So I might say let $a = 1$ and $b = 1$

then $a = b$ and $a^2 = b^2$ thus $a^2 - b^2 = 0$

Now that seems fine but I'm given the complex number $(a - ib)^2 = 4i$

Now I know $i^2 = -1$

So here's my attempt

$(a - ib)(a - ib) = 4i$

$a^2 - abi - abi + (bi)^2 - 4i = 0$

$a^2 - 2abi - b^2 - 4i = 0$

$ a^2 - b^2 -(ab + 2)2i = 0$

I've obviously not grasped this correctly as this certainly is not what I have been asked to prove.

Please could I have come guidance on how to solve this simple equation.

I noticed I left out what I am intending to prove which has been included in the title now namely:

Prove that $a^2 - b^2 = 0$

Thanks!

Metamorphosis
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6 Answers6

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You're doing great. Two complex numbers are equal when the real and imaginary parts agree. Hence you need to solve the system of equations $$\{a^2-b^2=0,~~ -(ab+2)2=0\}$$

vadim123
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  • Thanks for your input. I left out what I've been asked to prove though. The question further states "Prove that a^2 - b^2 = 0" Do I continue to solve for the aforementioned parts? – Metamorphosis Oct 24 '15 at 17:45
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You're on the right track: If $u + iv = 0$ for real numbers $u, v$, we must have $u = v = 0$. So, your equation $$a^2 - b^2 - (a b + 2) 2 i$$ is equivalent to the system $$ \left\{ \begin{array}{rcl} 0 &=& a^2 - b^2 \\ 0 &=& 2(a b + 2) \end{array} \right. $$

To solve this system, note first the we can rewrite the first equation as $$0 = (a - b) (a + b),$$ so $b = \pm a$.

Travis Willse
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You are on the right track, but you are overthinking it. The basic idea behind these sorts of questions is the fact that if $z = x + iy$ and $w = q + ip$ where $z,w \in \mathbb{C}$ and $x,y,q,p \in \mathbb{R}$ then $z = w$ iff $x = q$ and $y = p$.

Equipped with this we may deal with your problem. Indeed first we compute the left side $$ (a - ib)^2 = (a^2 - b^2) + i (-2ab) = x + iy$$ and $$ 4i = 0 + 4i = p + i q$$ Now we must have $$x = p \iff a^2 - b^2 = 0, \ y = q \iff 4 = -2ab$$ Using these two equations can you now solve for $a$ and $b$?

Kayle of the Creeks
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you should use $(x+y)^2=x^2+y^2+2xy$

$(a-ib)^2=4i$

$a^2+i^2b^2-2iab=4i$

$(a^2-b^2)-2iab=0+4i$ compare real and imaginary parts of both sides

$a^2-b^2=0\ \ ........(1)$

$ab=-2\ \ ........(2)$

Bhaskara-III
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We could use the general approach for solving $z^n = Re^{i\theta}$

First, let $z$ = $a-bi$

Then we have $$z^2 = 4i = 4e^{({\pi \over 2})i} = 4e^{({\pi \over 2}+2k\pi)i}, k = 0, 1$$

$$z = 2e^{({\pi \over 4}+{k\pi)i}}, k=0,1$$

You can then convert $z$ back to cartesian form and compare with $a-bi$.

Nicholas
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Since $\sqrt{i} =\pm\frac{1+i}{\sqrt{2}} $, $a-ib =\pm2\frac{1+i}{\sqrt{2}} =\pm(1+i)\sqrt{2} $, so $a=\pm\sqrt{2}$ and $b=\mp\sqrt{2}$.

marty cohen
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  • There is no need to introduce the ill-defined notion $\sqrt{i}$ to solve this question. – Did Oct 24 '15 at 17:48
  • There is no need to solve this question. – marty cohen Oct 24 '15 at 18:33
  • Huh? $ $ $ $ $ $ – Did Oct 24 '15 at 19:49
  • Just being snarky, Introducing $\sqrt{i}$ makes solving the problem easy, so I don't understand your phrase "There is no need to introduce the ill-defined notion". – marty cohen Oct 24 '15 at 21:47
  • Here is an explanation of this phrase: first, this notion is ill-defined; second, it is not needed to solve the question. No, introducing $\sqrt{i}$ does not make easy to solve the problem, only easier to make mistakes while trying to solve it. By the way, why should $\sqrt{i}$ be a two-elements set but not $\sqrt{2}$? – Did Oct 24 '15 at 21:53
  • Then use $|\sqrt{2}|$ if that will make you happier. My point is that there are two possible square roots of $i$, we can write them explicitly, and doing this makes solving the problem easy. – marty cohen Oct 24 '15 at 22:27
  • Sorry but do you know many people who use $\sqrt{2}$ to denote the pair ${-1.141...,+1.141...}$? Probably not, which indicates why using $\sqrt{i}$ in general is not a good idea. Furthermore, in the present case, there are proofs which avoid this, and they are better. – Did Oct 24 '15 at 22:35
  • I was just eliminating your stated ambiguity. And I do not care if you think that they did better. – marty cohen Oct 24 '15 at 22:38