0
$\begingroup$

Find all the complex solutions to the equation $$iz^2+(3-i)z-(1+2i)=0$$

I've tried to solve this equation with two different approaches but in both cases I couldn't arrive to anything.

1) If $P(z)=iz^2+(3-i)z-(1+2i)$, then the original problem is equivalent to finding the roots of $P$. If I consider the polynomial $P(z)\overline{P}(z)$, then this polynomial has integer coefficients, which I thought it could make things a little bit easier.

So $P(z)\overline{P}(z)=x^4-3x^3+6x^2-10x+5$, I didn't know what to do after this.

2) I've tried to solve the equation directly, $$iz^2+(3-i)z-(1+2i)=0$$ if and only if $$z(iz+3-i)=1+2i$$

The number of the left equals the number of the right if and only if they have the same module and argument. At this point I got stuck trying to actually calculate the argument and modulus of the product on the left side of the equation.

I would appreciate any help, thanks in advance.

$\endgroup$
1
  • $\begingroup$ FYI, the quadratic formula from the reals works for any field not of characteristic 2 $\endgroup$
    – Alan
    Sep 3, 2015 at 3:39

7 Answers 7

7
$\begingroup$

Hint: It is a quadratic equation with $a=i$, $b=3-i$, and $c=-(1+2i)$.

$\endgroup$
5
$\begingroup$

Its roots(let $\alpha,\beta$) are given by quadratic formula.

$\alpha,\beta=\frac{i-3\pm\sqrt{i^2+9-6i+4i(1+2i)}}{2i}=\frac{i-3\pm i\sqrt2 \sqrt i}{2i}=\frac{i-3\pm i\sqrt2(\frac{1+i}{\sqrt2})}{2i}=1+2i,i$

$\endgroup$
2
$\begingroup$

Since the quadratic equation has complex coefficients, it is not the case that the two solutions are complex conjugates of one another. However, the two solutions do lie diametrically opposite one another on a circle of radius $ \ \rho \ $ centered at a complex number $ \ z_0 \ $ . The diameter makes an angle $ \ \theta \ $ to the "real" axis, so the two solutions can be expressed as $ \ z \ = \ z_0 \ \pm \ \rho \ cis \ \theta \ $ or $ \ z \ = \ z_0 \ \pm \ \rho \ e^{ \ i \theta} \ $ . (I am expressing it this way to demonstrate how this might be done if you aren't used to taking square-roots of complex numbers.)

If we work with a quadratic polynomial with a leading coefficient of 1 , it may be factored as $ \ ( z \ - \ z_0 \ - \ \rho \ e^{ \ i \theta} ) \ ( z \ - \ z_0 \ + \ \rho \ e^{ \ i \theta} ) \ $ . Multiplying this out for the quadratic equation produces

$$ z^2 \ - \ 2 \ z_0 \ z \ + \ ( \ z_0^2 \ - \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ ) \ = \ 0 \ \ . $$

Martin Sleziak shows the monic polynomial equation we wish to solve, so we can compare coefficients with $ \ z^2 \ -(1+3i) \ z \ + \ (-2+i) \ = \ 0 \ $ . We have immediately that $ \ z_0 \ = \ \frac{1 \ + \ 3i}{2} \ $ . [In fact, what we are doing is essentially what "completing the square" provides us.] The square of this number is $ \ z_0^2 \ = \ -2 \ + \ \frac{3}{2}i \ $ , allowing us to establish the equation

$$ z_0^2 \ - \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ = \ -2 \ + \ i $$ $$ \Rightarrow \ \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ = \ ( -2 \ + \ \frac{3}{2}i ) \ - \ ( -2 \ + \ i ) \ = \ \frac{1}{2}i \ \ . $$

The modulus of $ \ e^{ \ i \cdot 2 \theta} \ $ is 1 , so we can write $ \ | \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ | \ = \ | \ \rho^2 \ | \ = \ | \ \frac{1}{2}i \ | \ = \ \frac{1}{2} \ $ . Inserting this in the earlier equation leaves us with $ \ \frac{1}{2} \ \cdot \ \ e^{ \ i \cdot 2 \theta} \ = \ \frac{1}{2}i \ \ \Rightarrow \ \ e^{ \ i \cdot 2 \theta} \ = \ i \ $ .

Now we have reached the part where we are extracting the square-root. "Euler's Identity" gives us $ \ e^{ \ i \cdot 2 \theta} \ = \ \cos \ 2 \theta \ + \ i \ \sin \ 2\theta \ = \ 0 \ + \ 1 \cdot i \ $ . It remains to solve the simultaneous trigonometric equations

$$ \cos \ 2 \theta \ = \ 0 \ \ \text{and} \ \ \sin \ 2 \theta \ = \ 1 \ \ \Rightarrow \ \ 2 \theta \ = \ \frac{\pi}{2} \ + \ 2k \cdot \pi \ \ \Rightarrow \ \ \theta \ = \ \frac{\pi}{4} \ + \ k \cdot \pi \ \ . $$

So the two angles involved are $ \ \frac{\pi}{4} \ \ \text{and} \ \ \frac{5 \pi}{4} \ $ (and, redundantly for our purpose, all the other "angle-names" in these directions). Bringing together our results, the two solutions to the complex quadratic equation are

$$ \frac{1 \ + \ 3i}{2} \ + \ \rho \ (\cos \ \frac{\pi}{4} \ + \ i \ \sin \ \frac{\pi}{4} ) \ \ , \ \ \frac{1 \ + \ 3i}{2} \ + \ \rho \ (\cos \ \frac{5 \pi}{4} \ + \ i \ \sin \ \frac{5 \pi}{4} ) $$ $$ = \ \frac{1 \ + \ 3i}{2} \ + \ \frac{1}{\sqrt{2}} \ ( \frac{\sqrt{2}}{2} \ + \ i \ \frac{\sqrt{2}}{2} ) \ \ , \ \ \frac{1 \ + \ 3i}{2} \ + \ \frac{1}{\sqrt{2}} \ ( -\frac{\sqrt{2}}{2} \ - \ i \ \frac{\sqrt{2}}{2} ) $$ $$ = \ \left(\frac{1 }{2} \ + \ \frac{1}{ 2 } \right) \ + \ i \ \left( \frac{3}{2} \ + \ \frac{1}{2} \right) \ \ , \ \ \left(\frac{1 }{2} \ - \ \frac{1}{ 2 } \right) \ + \ i \ \left( \frac{3}{2} \ - \ \frac{1}{2} \right) $$

$$ = \ \ 1 \ + \ 2i \ \ , \ \ 0 \ + \ i \ \ , $$

agreeing with the solutions shown by Vinod Kumar Punia and by WA. (I will say that things work out rather tidily for this equation; we would have somewhat more work to do in general to solve the pair of trigonometric equations, but it can still be managed "by hand" with the real and imaginary parts of the coefficients being integers or even rational numbers.)

$\endgroup$
3
  • $\begingroup$ Thanks for your detailed answer though I couldn't understand one thing: why must the two solutions lie diametrically opposite one another on a circle? $\endgroup$
    – user16924
    Sep 5, 2015 at 1:41
  • 1
    $\begingroup$ Note that upon completing-the-square or using the quadratic formula, we get [number] ± √(negative number) . The first number corresponds to the complex number that marks the center of the circle. The square root gives a complex number which is being added or subtracted vectorially to the first one (complex numbers add like vectors on the complex plane). So we have a vector or its opposite (a vector of the same length with opposite direction) being added to the first vector. The vector sums are the two solutions to the complex-coefficient quadratic equation (continued) $\endgroup$ Sep 5, 2015 at 2:18
  • 1
    $\begingroup$ and can be thought of as lying on the diameter of a circle [note that the angles we found are $ \ \pi \ $ radians (180º) apart]. We have the same thing for a real-coefficient quadratic equation with two solutions: $ \ -\frac{b}{2a} \ $ is the center of an interval on the real number line and $ \ \frac{\sqrt{b^2 - 4ac}}{2a} \ $ is the distance from the center at which the two real solutions lie. $\endgroup$ Sep 5, 2015 at 2:21
2
$\begingroup$

As already mentioned in other comments and answers, quadratic equations are solved in $\mathbb C$ in a similar way as in $\mathbb R$. So you can use the quadratic formula. The main difference is that this time the square root of a complex number $D$ means: "substitute here any solution of the equation $x^2=D$".

If, for some reason, you are not comfortable with using square root of a complex number, you can use another method which works for real numbers - completing the square. (Which is how quadratic formula can be derived.) You get \begin{align} iz^2+(3-i)z-(1+2i)&=0\\ z^2-(1+3i)z+(-2+i)&=0\\ \left(z-\frac{1+3i}2\right)^2&=\left(\frac{1+3i}2\right)^2+2-i \end{align} Now it only remains to simplify the RHS and find the complex square roots of the RHS.

$\endgroup$
2
$\begingroup$

$$iz^2+(3-i)z-(1+2i)=0 \Longleftrightarrow$$ $$\left(1+iz\right)\left(z-(1+2i)\right)=0 \Longleftrightarrow$$ $$i\left((-2+i)+(-1-3i)z+z^2\right)=0 \Longleftrightarrow$$ $$(-2+i)+(-1-3i)z+z^2=0 \Longleftrightarrow$$ $$\left(z+(-1-2i)\right)\left(-i+z\right)=0 \Longleftrightarrow$$ $$\left(z+(-1-2i)\right)=0 \Longleftrightarrow \vee \left(-i+z\right)=0 \Longleftrightarrow$$ $$z+(-1-2i)=0 \Longleftrightarrow \vee -i+z=0 \Longleftrightarrow$$ $$z=1+2i \vee z=i $$

$\endgroup$
1
  • $\begingroup$ Yes, it was a bit embarrassing to be working out the "machinery" for the complex square roots and discover along the way that if one had simply thought to test $ \ i \ $ in the given equation, one could have immediately factored the whole thing... $\endgroup$ Sep 5, 2015 at 3:52
1
$\begingroup$

Perhaps you could make it a real quadratic equation by multiplying out by $i$. $-z^2 + (3i + 1)z + 2 - i = 0$, that makes, $-z^2 + (3z - 1)i + z + 2 = 0$, giving $\frac{z^2 - z - 2}{3z-1} = i$.

then you could square this and solve for $z$

$\endgroup$
0
$\begingroup$

Use the AC method, just with complex numbers!

Multiply by $i$ to get $(iz)^2 + (3-i)(iz) - (i - 2) = 0$. Now let $u = iz$. Then we have $u^2 + (3-i)u - i+2 = 0$, which can be factored as $(u + 1)(u + (2-i)) = 0$. Thus $u = -1, -2+i$ and since $u=iz$, $z= \frac{-1}{i}, \frac{-2+i}{i} = i, 2i+1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .