how to find a square root of a complex number?

Question

I'm confused of what $\sqrt {3 + 4i}$ would be after I used quadratic formula to simplify $z^2 + iz - (1 + i)$

Answer 1

Recall that

$$z=x+iy=|z|(\cos \theta+i\sin \theta)$$$$\implies \sqrt z=\sqrt{|z|}\left(\cos \left(\frac{\theta}2+k\pi\right)+i\sin \left(\frac{\theta}2+k\pi\right)\right),\,k=0,1$$

Answer 2

Well. As a hint: a complex number can be represented by a real part and an imaginary part. Or, on the complex plane, it can be expressed as a distance from the origin (its magnitude) and an angle. Multiplying two complex numbers multiplies the magnitudes and adds the angles—and as with real numbers, there are two square roots.

Gimusi's answer uses that approach.

Also I think I might detect a $3,4,5$ triangle somewhere in there . . .

Answer 3

Let $\sqrt{3+4i}=x+yi$, with $x,y\in\Bbb R$. Then $3+4i=(x+yi)^2=x^2-y^2+2xyi$.

Equating real and imaginary parts gives $x^2-y^2=3$ and $2xy=4$. The second equation gives $y=2/x$, and substituting this back into the first equation gives $x^2-4/x^2=3$.

Multiplying through by $x^2$ gives a quadratic in $x^2$, namely $x^4-3x^2-4=0$. This factorises as $(x^2-4)(x^2+1)=0$. $x$ is real, so we can't have $x^2+1=0$. Hence $x^2-4=0$, i.e. $x=\pm 2$.

I will let you add the finishing touches.

Answer 4

If you want to understand where a formula for the real and imaginary parts of square roots comes from, set $(x+yi)^2 = a+bi$, where we know $a$ and $b$ we want to find $x$ and $y$. Assume $b \not= 0$ (otherwise $a + bi = a$ is a real number and you know what to do whether $a > 0$ or $a < 0$, or $a = 0$). Then $x^2 - y^2 = a$ and $2xy = b$. Since $b \not= 0$, $x$ and $y$ are nonzero and we can write $y = b/(2x)$. Then $x^2 - b^2/(4x^2) = a$, so $4x^4 - 4ax^2 - b^2 = 0$.

Actually, we could have derived that same equation even if $b = 0$: multiply $x^2 - y^2 = a$ on both sides by $x^2$ to get $x^4 - (xy)^2 = ax^2$, so $x^4 - b^2/4 = ax^2$. Multiplying through by $4$ and bringing all the terms to the left, we get $4x^4 - 4ax^2 - b^2 = 0$ again. Thus this quartic in $x$ is valid no matter what $b$ is.

The quartic in $x$ is quadratic in $x^2$, so $$ x^2 = \frac{4a \pm \sqrt{16a^2 + 16b^2}}{2\cdot 4} = \frac{a \pm \sqrt{a^2 + b^2}}{2}. $$ For $x$ to be real, $x^2$ needs to be positive so we must use the $+$ sign on the right. Thus $$ x^2 = \frac{a + \sqrt{a^2 + b^2}}{2} \geq 0, $$ so $x$ is the positive or negative square root of this real number. The roles of $x$ and $y$ are nearly symmetric: we can write $y^2 - x^2 = -a$ and $2yx = b$, so running through the same argument with $a$ changed to $-a$, we have $$ y^2 = \frac{-a + \sqrt{a^2 + b^2}}{2} \geq 0, $$ so $y$ is the positive or negative square root of this real number.

Watch out: we gave two values for $x$ and two for $y$, but those choices are not independent (otherwise we'd have four square roots of $a+bi$ in $\mathbf C$). They are linked by the relation $2xy = b$, and this linkage means the signs of $x$ and $y$ (as square roots) match if $b > 0$ and they are opposite if $b < 0$ (figure out what happens if $b = 0$).