Complex number equation $x^2+x+(1-i)=0$

Question

is there any easy way to solve the following equation: $$x^2+x+(1-i)=0$$ I have tried to write $-3+4i$ in the trigonometric way but I do not think that there is any normal way to get angle $\alpha$ such that $\cos(\alpha)=-3/5$ and $\sin(\alpha)=4/5$....

Answer 1

Having found the discriminant $-3+4i$, let $u+vi$ be a square root. Thereby, squaring both sides gives

$(u+vi)^2=(u^2-v^2)+(2uv)i=-3+4i$.

So

$2uv=4,v=2/u$

$u^2-v^2=u^2-(4/u^2)=-3,u^4+3u^2-4=0,(u^2)^2+3(u^2)-4=0$.

For $u$ to be real its square must be nonnegative. Solve the real-variable quadratic equation accordingly for $u^2$, take the square roots for $u$ and then back-substitute $v=2/u$.

Answer 2

For this particular quadratic equation, there is a very easy way to solve it: simple inspection suggests $x=i$ as a root of $x^2+x+(1-i)$, which is, indeed, easy to verify:

$$i^2+i+(1-i)=-1+i+(1-i)=0$$

The other root is easy to find from the fact that the sum of the roots of a quadratic is the negative of the coefficient of $x$, in this case $-1$, so the other root is $-1-i$. (As a doublecheck, note that the product of the roots, $i(-1-i)$, is equal to the constant term, $1-i$.)

To be a little fancy about what we've just done, the quadratic $x^2+x+(1-i)$ is a monic polynomial over the Gaussian integers, with prime constant coefficient $1-i$, so if it factors over the Gaussian integers, one of the roots must be a unit, i.e., one of $1$, $-1$, $i$, or $-i$.

Answer 3

Factorize the equation as

$$x^2+x+(1-i)=x^2+x-i(1+i)=(x-i)(x+1+i)=0 $$

which yields the solutions $x=i,\>-(1+i)$.

Answer 4

Apparently you want to use the quadratic formula and are puzzled by the need to find the square root of $-3+4i$. The other answerers want to circumvent that question (which is fine to an extent). Let me show how to find that square root using the half-angle formulas.

We have $$ z=-3+4i=r(\cos\theta+i\sin\theta), $$ and easily calculate that the modulus $r=5$. We also observe that the number $z$ is in the second quadrant, so we can assume that $\theta\in(\pi/2,\pi)$. This implies that $\theta/2$ is in the first quadrant. Furhtermore, $5\cos\theta=-3$, so we can deduce that $\cos\theta=-3/5$.

The half-angle formulas then give that trig functions at $\theta/2$ are (both positive, because we saw that $\theta/2\in(0,\pi/2)$) $$ \begin{aligned} \cos\frac\theta2&=\sqrt{\frac{1+(-3/5)}2}=\sqrt{\frac15}=\frac1{\sqrt5},\\ \sin\frac\theta2&=\sqrt{\frac{1-(-3/5)}2}=\sqrt{\frac45}=\frac2{\sqrt5}. \end{aligned} $$ Therefore, for this choice of branch of the complex square root $$ \sqrt{z}=\sqrt{5}(\cos\frac\theta2+i\sin\frac\theta2)=1+2i. $$ It is, of course, easy to check this by squaring: $(1+2i)^2=-3+4i$.

Moral: You don't need the angle nor the half-angle. You only need their co/sines.

Answer 5

Well $x = \frac {-1 \pm \sqrt{1-4(1-i)}}2=\frac {-1\pm \sqrt{-3+ 4i}}2$

So what is $\sqrt{-3 + 4i}$.

Well converting to polar coordinates $-3+4i = 5*e^{i\alpha}$ where $\alpha$ is the angle where $\cos \alpha -3$ and $\sin \alpha = 4$. So to find that we have $\alpha = \arctan \frac {\sin \alpha}{\cos \alpha}=\arctan (-\frac 43)$.

So $\sqrt{-3 +4i} = \sqrt5e^{\pm i\arctan(-\frac 43)}$

And $x = \frac {-1+\sqrt5e^{\pm i\arctan(-\frac 43)}}2=$

$\frac {-1\pm \sqrt5\cos \arctan(-\frac 43)}2 \pm i\frac {\sqrt 5\sin \arctan (-\frac 43)}2$

That top term $\frac {-1+\sqrt 5e^{\pm i\arctan(-\frac 43)}}2$ seems to me a legitimate answer although we have not sense of word that complex number is "like".

(actually we can put that in terms of $re^{i\phi}$ by the half angle theorem.... But... someone else answered how to do that. I ..... don't want to... never liked trig identities mumble mumble.)

The second $\frac {-1\pm \sqrt 5\cos \arctan(-\frac 43)}2 \pm i\frac {\sin \arctan (-\frac 43)}2$ doesn't seem kosher to me as arctrig function of trig functions seem like we are just shunting the issue away.

Which we can face head on with trig and the half angle theorem.... But I won't 'cause I'm lazy and someone else did.....

But to do it algebraically... (which I'm sure the book intended in the first place.)

$\sqrt{-3+4i} = a + bi$ (for real $a,b$)

then $a^2 - b^2 = -3$ and $2ab = 4$. So $a = 2\frac 1b$ so $4*\frac 1b^2 - b^2 =-3$ so $4 -b^4 =-3b^2$ or $b^4 -3b^2 -4=0$ so $(b^2+1)(b^2 - 4) =0$. As $b$ is real $b^2 =4$ and $b=\pm 2$ and $a=\pm1$.

So $\sqrt{-3+4i} = \pm (1+2i)$. And $(1+2i)^2 = (1-4) + 2*2i = -3 +4i$.

So $x = \frac {- 1 \pm (1+2i)}2 = -1-i$ or $i$

Let's check:

$x= i$ then $x^2 + x + (1-i) = -1 + i + (1-i) = 0$.

And $x = -1-i$ then $x^2 + x +(1-i) = (0 +2i) + (-1-i) +(1-i) = 0$.

Yup they work.

Answer 6

We may also be able to dodge having to find the square-root of a complex number by making use of the properties of the zeroes of this polynomial. If we call the zeroes $ \ r \ $ and $ \ s \ \ , $ then their sum is $$ r \ + \ s \ \ = \ \ -\frac{b}{a} \ \ = \ \ -\frac11 \ \ = \ \ -1 \ \ . $$ If we express the complex zeroes as $ \ r \ = \ \alpha + \beta·i \ $ and $ \ s \ = \ \gamma + \delta·i \ \ , $ the result for the sum tells us that the zeroes has opposite imaginary parts and we may write $ \ r \ = \ \alpha + \beta·i \ $ and $ \ s \ = \ (-1-\alpha) - \beta·i \ \ , $

The product of the zeroes is then $$ \ r · s \ \ = \ \ \frac{c}{a} \ \ = \ \ [ \ (\beta^2 - \alpha^2 - \alpha) \ + \ i·(-\beta)·(2·\alpha + 1) \ ] \ \ = \ \ 1 \ - \ i $$ $$ \Rightarrow \ \ \beta^2 - \alpha^2 - \alpha \ \ = \ \ 1 \ \ \ , \ \ \ \beta·(2·\alpha + 1) \ \ = \ \ 1 \ \ . \quad \mathbf{[ \ 1 \ ]} $$

Alternatively, the difference of the zeroes is $ \ r \ - \ s \ = \ \frac{\sqrt{\Delta}}{a} \ \ ; $ to avoid needing to find the square-root, we will square this difference to obtain $$ (r \ - \ s)^2 \ \ = \ \ \Delta \ \ = \ \ -3 \ + \ 4i \ \ = \ \ [ \ (1 + 2·\alpha) \ + \ i·2·\beta \ ]^2 $$ $$ \Rightarrow \ \ 4·(\alpha^2 \ + \ \alpha \ - \ \beta^2) \ + \ 1 \ \ = \ \ -3 \ \ \ , \ \ \ 4·i·(2·\alpha·\beta \ + \ \beta) \ \ = \ \ 4i \ \ , $$

which equations are equivalent to those in $ \ \mathbf{[1]} \ $ above.

It is fairly easy here to find the solution $ \ \alpha \ = \ 0 \ \ , \ \ \beta \ = \ 1 \ \ $ by inspection. Thus, the zeroes of the quadratic equation are $ \ \mathbf{r \ = \ i \ \ , \ \ s \ = \ -1 - i } \ \ . $