Square Roots of Complex Number $3-4i$

Question

What I did

$z^2=3-4i$

$(a+bi)^2 = 3-4i$

$a^2-b^2+2abi = 3-4i$

Then got 2 simultaneous equations

$a^2-b^2=3$ and $2ab=-4$

Solve for $a^2$ in 1st equation: $a^2=3+b^2$

Subbed into 2nd equation to power of 2

$(3+b^2)b^2=4$

$b^4+3b^2-4=0$

Is there a better way than below? Solving power 4 equation then cubic?

Then solved solved power 4 equation ...

$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$

$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$

then solved for cubic equation, getting ...

$(b-1)(b+1)(b^2+4)=0$

So

$b=\pm 1 \text{ or } 2i$

What did the book do now?

I did: (subbing into $a^2=3+b^2$)

When $b=1, a^2=3+1^2 \Longrightarrow a = 2$

When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$

When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$

So I will have 3 equations

• $2+i$ // why is this not in the book?
• $2-i$
• $i-2$

UPDATE (In response to @David Mitra)

Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$

Then when $y=1, x=4$. $y=-4, x=-1$

Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$

Finally testing equations:

$(2+i)^2 = 3+4i$ (rej)

$(2-i)^2=5$ what to do?

$(i-2)^2=3$

$(-2-i)^2=3$

... 8 equations (but 4 unique)

Answer 1

A better way to solve $ b^4+ 3b^2-4=0 $ is to write it as $(b^2 +4)(b^2 -1)=0$. This gives $b=1,-1,2i,-2i$. (Note, you missed $-2i$ in your write up.)

Then find the $a$ values as you did...

As for why $2+i$ wasn't in the answer key:

When you squared both sides of $ab=-2$ to get $a^2b^2=4$, you introduced new solutions. So in the end, you need to check which solutions you obtained work. This is why $2+i$ wasn't a solution to the original problem. In fact: $(2+i)^2=3+4i$.

In your last step, when solving for $a$ when $b=2i$, you obtain $a^2=-1$, so $a=i$ or $a=-i$ (you missed $-i$).

For $b=-2i$, you get again $a=i$ or $a=-i$.

This gives you four solutions that need checking: $-2+i, -2-i, 2+i, 2-i$.

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Your update is off...

You are correct up to the point when you obtain $b^4+3b^2-4=0$. As I mentioned above, to solve this, write $$ b^4+ 3b^2-4=0 \iff (b^2 +4)(b^2 -1)=0\iff b=1,-1,2i,-2i. $$ For each of these $b$ values, we find the corresponding $a$ value(s):

For $b=1$: $a^2=3+1^2\Rightarrow a=2,-2$.

This gives $a+bi=2+i$

or $a+bi=-2+i$.

For $b=-1$: $a^2=3+(-1)^2\Rightarrow a=2,-2$.

This gives $a+bi=2-i$

or $a+bi=-2-i$.

For $b=2i$: $a^2=3+(2i)^2\Rightarrow a^2=-1\Rightarrow a=i,-i$.

This gives $a+bi=i+(2i)i=-2+i$

or $a+bi=-i+(2i)i=-2-i$.

For $b=-2i$: $a^2=3+(-2i)^2\Rightarrow a^2=-1\Rightarrow a=i,-i$.

This gives $a+bi=i+(-2i)i=2+i$

or $a+bi=-i+(-2i)i=2-i$.

So we have four possible solutions: $$ 2+i, 2-i, -2+i , -2-i $$

Now we check which of these is a solution to the original equation $(a+bi)^2=3-4i$. Taking the squares of the possible solutions: $$(−2+i)^2=3−4i, (−2−i)^2=3+4i, (2+i)^2=3+4i, (2−i)^2=3−4i.$$

So, there are two solutions: −2+i and 2−i.

Answer 2

You can look at this as a problem in the arithmetic of the Gaussian Integers, $\mathbb Z[i]$. I’ll make use of the fact that this ring is a Unique Factorization Domain, and I’ll also use the fact that every prime $p\equiv1\pmod4$ is writable as the sum of two squares, $p=m^2+n^2$, equivalently, $p=(m+ni)(m-ni)$, and that $m\pm ni$ are primes in $\mathbb Z[i]$.

For the prime $5$, the first where these facts become operative, we have $5=1^2+2^2=(1+2i)(1-2i)$, and certainly the two factors are not related by a unit, so that they generate different prime ideals of the G-Integers. What about $3+4i$? We have $(3+4i)(3-4i)=25=(1+2i)^2(1-2i)^2$, a product of primes in the UFD $\mathbb Z[i]$. Up to units, you see that it must be that $3+4i$ is equal to $(1+2i)^2$ or $(1-2i)^2$. You check, and you see that you don’t need to worry about units, and the first possibility is the right one.

Answer 3

The value of $a$ is calculated incorrectly when $b=1$. From $2ab=-4$ there is a unique value of $a$ for every value of $b$. When $b=1$, $a$ can only have the value $-2$ (not $+2$ as stated in the question), and then $z = i - 2$.

The quartic equation for $b$, although derived by squaring, is correct and is not responsible for introducing any extra solutions. It could have been derived without squaring, using $ab=−2$. The error is in the step "subbing into $a^2=3+b^2$ ". That quadratic equation derives two values of $a$ from each value of $b$, when we know from $2ab=−4$ that there is only one value of $a$ for each value of $b$. If $2ab=-4$ had been used for the determination of $a$ from $b$ no extra solutions would appear, and every value of $b$ would determine a correct value of $a$.

What did the book do now?

The book does not say, but I guess it meant: eliminate one of the variables using $xy = -2$, obtaining a quartic (degree 4) equation in the other variable that is a quadratic equation in $x^2$ (or $y^2$, whichever is the surviving variable). The quartic has two real and two imaginary roots. Where the book says there are two solutions, that means solutions with real values of $x$ and $y$. These can be found by taking the two real solutions of the quartic and solving for the other variable through $xy=-2$; if one variable is real, the value found for the other variable will also be real.

Answer 4

$b^4+3b^2-4=0$ is a quadratic equation in $b^2$, i.e. you can write $c=b^2$, and then the equation says $c^2+3c-4=0$. If you know how to solve quadratic equations, you've got it.

Answer 5

The high school way:

If $(x+iy)^2=3-4i$, expanding the l.h.s. and identifying the real and imaginary parts, you obtain the system of equations $\;\begin{cases} x^2-y^2=3,\\2xy=-4. \end{cases}$

Now, you can also write that $|x+iy|^2=|3-4i|$, i.e. $\;x^2+y^2=5$. Therefore you obtain the linear system in $x^2$ and $y^2$: $$\begin{cases} x^2-y^2=3,\\x^2+y^2=5 \end{cases}\iff \begin{cases} x^2=4,\\y^2=1 \end{cases}$$ Thus, $x=\pm 2,\;y=\pm 1$. Furthermore, the unused equation $xy=-2$ says that $x$ and $y$ have opposite signs. Consequently the solutions are $$x+iy=\pm(2-i).$$