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Super basic question but some reason either I'm not doing this right or something is wrong.

The best route usually with these questions is to transform $3+4i$ to $re^{it}$ representation.

Ok, so $r^2=3^2+4^2 = 25$, so $r=5$. And $\frac{4}{3}=\tan(t)$ so that means $t \approx 0.3217$ and I'm not going to get an exact answer like that.

Another method would be to solve quadratic formula $z^2-3-4i=0$ that means $z_0=\frac{\sqrt{12+16i}}{2}$ and $z_1=\frac{-\sqrt{12+16i}}{2}$

But now I have the same problem, $12+16i$ doesn't have a "pretty" polar representation so its difficult to find $\sqrt{12+16i}$

I want to find an exact solution, not approximate, and it should be easy since the answers are $2+i$ and $-2-i$

Edit:

Also, something else is weird here. I know that if $z_0$ is some root of a polynomial then it's conjugate is also a root.but $2+i$ and $-2-i$ are not conjugates.

Git Gud
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Oria Gruber
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  • Try $(x+iy)^2=3+4i$. Regarding the edit, that's for real polynomials (recall the proof of that result). – Git Gud Feb 02 '15 at 10:53
  • yeah I just tried the $(x+iy)^2$ method after I finished writing this question, it works. But what if the question was to find $z$ that satisfies $z^{20}=3+4i$? It's not always a viable method – Oria Gruber Feb 02 '15 at 10:54
  • You can use the top of this answer and the bottom of this one, but you're only getting pretty numbers as much as $\arctan$ yields pretty numbers. – Git Gud Feb 02 '15 at 10:56
  • @OriaGruber About your last lines: the complex roots of a real polynomial indeed appear as conjugate pairs, yet in this case you have the roots of a complex non-real polynomial, namely $;z^2-(3+4i)=0;$ – Timbuc Feb 02 '15 at 11:04

4 Answers4

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Hint:

$$z^2=3+4i=5e^{it+2k\pi i}\;,\;\;k\in\Bbb Z\;,\;\;t=\arctan\frac43\implies$$

$$z=\sqrt[2]5e^{\frac{it+2k\pi i}2}\;,\;\;k=0,1\;\;\text{(Why it is enough to take only these vales of}\;\;k\;?)$$

A more basic approach: put$\;z=a+bi\;,\;\;a,b\in\Bbb R\;$ , so that

$$3+4i=(a+bi)^2=(a^2-b^2)+2abi\implies\begin{cases}a^2-b^2=3\\{}\\2ab=4\implies b=\frac2a\end{cases}\;\;\implies$$

$$a^2-\frac4{a^2}=3\implies 0=a^4-3a^2-4=(a^2-4)(a^2+1)\implies a=\pm2$$

and thus

$$\;b=\pm\frac22=\pm1\;\implies a+bi=\begin{cases}\;\;\;2+i\\{}\\-2-i\end{cases}$$

Timbuc
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0

This can be computed by a Simple Square Root Denesting Rule:

$\ 3+4\sqrt{-1}\ $ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = -5\,\ $ yields $\,\ 8+4\sqrt{-1}\:$

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{16}\, =\, 4.\ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\ \ 2+\sqrt{-1}$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.

Community
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Bill Dubuque
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  • I think I understand "norm"=$|z|$ , but what do you substract it from? The number itself? Then you have gotten $-2+4\sqrt{-1};$, right this? And what is "trace"? Thank you –  Feb 02 '15 at 17:47
  • @Antoine Yes, we subtract out from $,z,$ the value $\sqrt{{\rm norm}(z)}, =, \pm |z|, =, \pm 5.,$ The rule works for either choice of sign of the sqrt. Here the arithmetic is simpler if we use $,-5.,$ Follow the links for more. – Bill Dubuque Feb 02 '15 at 17:58
  • Thank you, it's clear yet. Very nice sir. –  Feb 02 '15 at 18:07
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$$ z^2=3+4i\\ (a+bi)^2=3+4i,\;\;\;\;a,b\in\mathbb{R}\\ a^2+2abi-b^2=3+4i $$ Now you have a system of equations: $$ a^2-b^2=3\tag{1} $$ $$ 2ab=4\tag{2} $$ $$ b=\dfrac2a $$ Substitute $b$ in $(1)$: $$a^2-\dfrac4{a^2}=3$$ Now let $t=a^2$, then $$ t^2-3a-4=0\\ t=\dfrac{3\pm\sqrt{9+16}}{2} $$ which have solutions $t_1=4$ and $t_2=-1$. Notice that $t_2$ cannot be solution, so only solution is $$ t=4\\ a^2=4\\ a=\pm2 $$ So, we have solutions $a=2,b=1$ and $a=-2,b=-1$. Solutions of $z$ are $$z=2+i\;\;\;\;\text{or}\;\;\;\;z=-2-i$$

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We can use the double angle formula to find the components of $z$:

$$ \cos \theta = \pm \sqrt{\frac{1 + \cos 2\theta}{2}}$$

$$ \sin \theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}$$

Here $\theta$ is the argument of $z$ and $2\theta$ the argument of $z^2$.

We know $ \cos 2\theta = \frac{3}{5}$ so $\cos \theta = \pm \sqrt{\frac{4}{5}}$ and $\sin \theta = \pm\sqrt{\frac{1}{5}}$. Since $z^2$ lies on the first quadrant (between $0$ and $\pi/2$), its two square roots must lie on the first and third quadrants, respectively, therefore the final answer is

$$ z_1 = \sqrt{5} \left( \sqrt{\frac{4}{5}} + i\sqrt{\frac{1}{5}} \right) = 2 + i $$

$$ z_2 = -z_1 = -2 - i $$

Alternatively: Since the numbers are relatively simple, we can let $z = x + yi$. Then $$ z^2 = (x^2 - y^2) + 2xyi = 3 + 4i $$

Now equate the real and imaginary part to get this system of equations: $$ x^2 - y^2 = 3 $$ $$ 2xy = 4 $$

Dylan
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