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for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).

I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?

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  • $\begingroup$ Calculator? No kidding: there's no promise all angles will be "nice". Nevertheless, in this case you have that $\;\arctan\frac43=\theta\;$ and not the other way around. $\endgroup$ – Timbuc Feb 17 '15 at 9:47
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Note, we have $|w| = 5$. Let $\theta \in Arg(w)$ and then from your corresponding diagram of the triangle form my $w$, $\cos(\theta) = \frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$. Therefore, from $\sqrt{z} = \sqrt{z}\left( \cos(\frac{\theta}{2}) + i\sin(\frac{\theta}{2})\right )$, we essentially arrive at our answer.

Recall the half-angle identities of both cosine and sine. i.e.,

$$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1}{2}(1 + \cos(\theta))}$$

$$\sin \left (\frac{\theta}{2} \right) = \sqrt{\frac{1}{2}(1 - \cos(\theta))}$$

From plugging in the corresponding values into the above equations, we find that $\cos(\frac{\theta}{2}) = \frac{2}{\sqrt{5}}$ and $\sin(\frac{\theta}{2}) = \frac{1}{\sqrt{5}}$.

Then we obtain $\boxed{\sqrt{3 + 4i} = \pm (2 + i)}$

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Suppose $\sqrt{3+4i}$ were in standard form, say $x+yi$. Then we would have $$\begin{align} x+yi & = \sqrt{3+4i}\\ (x+yi)^2 & = 3+4i\\ (x^2-y^2) + 2xyi & = 3+4i \end{align} $$

so

$$\begin{align} x^2 -y^2 &= 3 \\ 2xy &= 4 \\ \end{align} $$

From the second equation we have $y = \frac2x$. Putting this into the first equation we obtain $$x^2 - \frac4{x^2} = 3.$$ Multiplying through by $x^2$, then setting $z=x^2$ we obtain the quadratic equation $$z^2 -3z -4 = 0$$ which we can easily solve to obtain $z=4$. (The other root, $z=-1$, is spurious since $z = x^2$ and $x$ is real.) Then since $x^2=z$ and $y=\frac2x$ we get $\color{darkblue}{x=2, y=1}$ and $\color{darkred}{x=-2, y=-1}$.

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Suppose you had $\theta = \tan^{-1} \frac34$. But you don't want $\theta$ itself; you want $x = r \cos \theta$ and $y = r\sin \theta$. The value of $\theta$ isn't required here; all you need are its sine and cosine.

This is fortunate because those are much easier to calculate than $\theta$ itself! In general, $\tan^{-1} \frac ab$ may be intractable, but even so, $\sin(\tan^{-1}\frac ab)$ and $\cos(\tan^{-1}\frac ab)$ are easy. Consider of this right triangle:

Right triangle with legs a, b

One sees immediately that since $\theta = \tan^{-1}\frac ab$, then $\sin(\tan^{-1} \frac ab) = \frac a{\sqrt{a^2+b^2}}$ and $\cos(\tan^{-1} \frac ab) = \frac b{\sqrt{a^2+b^2}}$.

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Were you told to find the square root of $3+4i$ by using Standard Form? This happens to be one of those situations where Pure Number Theory is more useful.

Your number is a Gaussian Integer, and the ring $\Bbb Z[i]$ of all such is well-known to be a Principal Ideal Domain. You find the factorization of a number like $3+4i$ by looking at its (field-theoretic) norm down to $\Bbb Q$: the norm of $a+bi$ is $(a+bi)(a-bi)=a^2+b^2$. Here the norm is $25$, so you’re confident that the only Gaussian primes dividing $3+4i$ are those dividing $25$, that is, those dividing $5$. But every prime congruent to $1$ modulo $4$ is the sum of two squares, and surenough, $5=4+1$, indicating that $5=(2+i)(2-i)$. The two factors there are (up to units $\pm1$, $\pm i$) the only factors of $5$, and thus the only possibilities for factors of $3+4i$. So you check:

Is $3+4i$ divisible by $2+i$, or by $2-i$? Do the division using high-school methods, and you see that it’s divisible by $2+i$, and wonderfully, the quotient is $2+i$. There you are, $\sqrt{3+4i\,}=2+i$, or its negative, of course.

But the moral of the story really is: if you’re going to work with Complex Numbers, you should play around with them computationally. If you had frolicked in the Gaussian world, you would have remembered the wonderful fact that $(2+i)^2=3+4i$, the point in the plane that gives you your familiar simplest example of a Pythagorean Triple.

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  • $\begingroup$ Very neat! I hope the poster of the question gives your answer a deep look. Though, I do not really know why your answer was downvoted. I assumed he/she was looking to put $\sqrt[]{3+4i}$ in Standard form. $\endgroup$ – Ozera Feb 17 '15 at 21:17
  • $\begingroup$ Maybe it was my error, @Ozera, to interject number theory into a question that almost surely arose in a complex-variable context. $\endgroup$ – Lubin Feb 17 '15 at 22:33
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you can do this without invoking the half angle formula explicitly. what you are after is $\cos(t/2)$ and $\sin t/2$ given $\cos t = \frac35$ and $\sin t = \frac45.$ let $O= (0,0), A = (1,0), B = (\frac35, \frac45)$ and $C$ be the midpoint of $AB.$ then $C$ has coordinates $(\frac45, \frac25).$ there are two points on the unit circle on the line $OC.$ they are $(\pm \frac2{\sqrt5}, \pm\frac{1}{\sqrt5}).$ since $\sqrt z$ has modulus $\sqrt 5,$ you get $\sqrt{ 3+ 4i }=\pm(2+i). $

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