Find all roots of $z^2=3-4i$.

Question

$z^2=3-4i$

$z^2+4i-3=0$

But how do I go on from here?

score 8 · Accepted Answer · answered Jan 21 '14 at 19:17

The way people will really solve it.

If we have a root $z_0$, then $-z_0$ is also a root. Since this is a quadratic, there can't be more than $2$ roots, so we just need to guess one root $z_0$. How do we do that?

Observe that $|z^2| = \sqrt{3^2 + 4^2} = 5$, therefore $|z|=\sqrt{5}$. Now, the only nice numbers with that absolute value are $\pm 1 \pm 2i$ and $\pm 2 \pm i$. Geometric intuition says that one of the roots must be found in the 4th quadrant. A quick check shows that indeed $z_0 = 2-i$ fits the bill: $$ (2-i)^2 = 4 - 4i + i^2 = 3 - 4i. $$

So the answer is $2-i$ and $-2 + i$.

This is how a lazy person like me solves a problem like this. It is perfectly legitimate to guess the answer (if you have a good argument why other answers are impossible, which in this case follows from the general knowledge that a quadratic has at most two roots). Only when guessing doesn't work right away do people start testing more powerful general approaches.

Hm, I'd really like to see the reasons for downvotes. Although be careful, I am this close to another rant on how smart guessing is the way things are actually done in mathematics. — Dan Shved, Jan 21 '14 at 19:25

score 4 · Answer 2 · answered Jan 21 '14 at 18:54

4

Hint: Write $3 - 4i$ in polar form,

$$3 - 4i = r e^{i\theta}$$

for an appropriate positive $r$ and $0 \le \theta < 2\pi$. Then use the fact that the two square roots of $re^{i\theta}$ are

$$\pm \sqrt{r} e^{i\theta/2}$$

answered Jan 21 '14 at 18:54

Are you sure about the $\pm$? This is usually taken into account by the fact that $r\mathrm{e}^{\mathrm{i}\theta} = r\mathrm{e}^{\mathrm{i}(\theta+2\pi n)}$ where $n$ is a whole number. – Fly by Night Jan 21 '14 at 18:56
@FlybyNight Yes, they'll give equivalent answers precisely due to the fact that $-e^{i\pi} = 1$. I didn't give the general method here since the question is only about square roots, and I don't think that the general method clarifies the solution in this case. – Jan 21 '14 at 19:00
1

IMHO, this answer is potentially misleading. The OP is quite likely to assume that the solution to $z^3 = 3-4\mathrm{i}$ is $\sqrt[3]{r}\mathrm{e}^{\mathrm{i}\theta/3}$ alone. This is a standard problem that asks for consideration modulo $2\pi$. – Fly by Night Jan 21 '14 at 19:03
Well the OP can consider $(se^{i\psi})^3 = re^{i\theta}$ so that $s = r^{\fract{1}{3}$ and $\phi = \fract{\theta}{3} + \fract{2pi*k}{3}$ – InsigMath Jan 22 '14 at 00:15
Sorry it should say $s = r^{\fract{1}{3}}$ – InsigMath Jan 22 '14 at 00:36

score 3 · Answer 3 · edited Apr 13 '17 at 12:20

Denesting $\,\sqrt{a+b\sqrt{n}}\,$ can be done by a simple rule that I discovered in my youth.

Simple Denesting Rule $\rm\ \ \, \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\:3-4i\:$ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -5\ $ yields $\ \ 8-4i\:$

which has $\rm\ \sqrt{trace}\: =\: \sqrt{16}\ =\ 4.\ \ \ \color{brown}{Dividing\ it\ out}\ $ yields $\ \ (8-4i)/4\, =\, 2-i$

Or: $\:3-4i\:$ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ \ = \ 5\ \ $ yields $\ {-}2-4i\:$

which has $\rm\ \sqrt{trace}\: =\: \sqrt{-4} = 2i.\ \ \color{brown}{Dividing\ it\ out}\ $ yields $\ (-2-4i)/(2i)\, =\, i-2$

One may choose the sign of the square-root so that the arithmetic is simpler, as in the first case above. For many further worked examples see my prior posts on denesting.

The guy who gave a downvote simply didn't get it. +1 – Patrick Da Silva Jan 22 '14 at 01:07 — Patrick Da Silva, Jan 22 '14 at 01:07

score 3 · Answer 4 · edited Apr 29 '14 at 02:54

3

Solve: $z^2 = 3 - 4i$

Let $$z= x + yi$$ Rewrite as $$(x + yi)^2 = 3 - 4i$$ Expand $$x^2 +2xyi + y^2i^2 = 3 - 4i$$ Simplify $$x^2 - y^2 + 2xyi = 3 - 4i$$ real/imaginary parts $$x^2 - y^2 = 3$$

and

$$2xy = -4$$ $$xy = -2$$ Therefore $y = -\frac{2}{x}$.

Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving $$x^2 - \frac{4}{x^2} = 3$$ Multiply by $x^2$: $x^4 - 4 = 3x^2$ Factor

$$x^4 - 3x^2 - 4 = 0$$ $$(x^2 + 1) (x^2 - 4) = 0$$ $$(x^2+ 1) (x + 2) (x - 2) = 0$$ Given $xy = -2$

for $x = 2$ then $y = -1$ therefore $z = 2 - i$

for $x = -2$ then $y = 1$ therefore $z = -2 + i$

edited Apr 29 '14 at 02:54

Anonymous Computer

5,639

answered Apr 28 '14 at 23:58

user146497

41

1

Please use complete sentences and better formatting so that your answer is as clear and understandable as possible. – Chris Brooks Apr 29 '14 at 00:42
This answer is simply unreadable. Recommending deletion. – Caleb Stanford Apr 29 '14 at 02:07
2

Edited answer. Please check if formatting is correct – Anonymous Computer Apr 29 '14 at 02:54

score 2 · Answer 5 · answered Jan 21 '14 at 18:54

2

HINT:

Let $\displaystyle z=a+ib$

So, we have $\displaystyle a^2-b^2+2ab i=3-4i$

Equate the real & the imaginary parts

Use $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$ to find $a,b$

Observe that the signs of $a,b$ are opposite

answered Jan 21 '14 at 18:54

lab bhattacharjee

274,582

I'm not sure where to go with the 3rd equation? – Mr Croutini Jan 21 '14 at 19:01
@MrCroutini, we have $a^2-b^2=3,2ab=-4,\implies a^2+b^2=\sqrt{3^2+(-4)^2}=+5$ Can you find $a^2,b^2$ form here? – lab bhattacharjee Jan 21 '14 at 19:03
No, sorry. I'm really not getting this.. :/ – Mr Croutini Jan 21 '14 at 19:08
@MrCroutini, add $a^2+b^2=5, a^2-b^2=3\implies 2a^2=8\implies a^2=4\iff a=\pm2$ and we have $2ab=-4\iff b=-\frac2a$ – lab bhattacharjee Jan 21 '14 at 19:09
Isn't $a^2-b^2=3^2-(-4)^2=-7$? – Mr Croutini Jan 21 '14 at 19:15
@lab This method can be packaged into a simple rule - see my answer. Happy denesting! – Bill Dubuque Jan 22 '14 at 00:04

Find all roots of $z^2=3-4i$.

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