complex numbers algebraically

Question

solve algebraically $x^2+2ix-5=3i$.

My Solution:

I tried using quadratic solution ( $a= 1 , b= 2i, c= -5-3i$ ) but it is wrong.

Answer 1

There is no escaping the issue of having to take the square-root of a complex number. "Completing the square" will also work (since it is the basis of the quadratic formula):

$$ z^2 \ + \ 2i \ z \ = \ 5 + 3i \ \ \Rightarrow \ \ z^2 \ + \ 2i \ z \ + \ i^2 \ = \ 5 + 3i + i^2 $$

$$ \Rightarrow \ \ (z + i)^2 \ = \ 4 + 3i \ \ \Rightarrow \ \ z \ = \ -i \ \pm \sqrt{4 +3i} \ \ . $$

There are the two complex-number solutions promised by the Fundamental Theorem of Algebra ("number of zeroes" version).

EDIT -- If you've had DeMoivre's Theorem and some trig, we can work this out a bit further to get exact values (though the numbers don't simplify). In polar form, $ \ 4 + 3i \ = \ 5 \ cis \ \theta \ , $ where $ \ \cos \theta = \frac{4}{5} \ , \ \sin \theta = \frac{3}{5} \ , $ and the "cis theta" notation is shorthand for $ \ \cos \theta \ + \ i \sin \theta \ . $ The square-roots we seek are

$$ (4 + 3i)^{1/2} \ = \ 5^{1/2} \ cis \frac{\theta}{2} \ , \ 5^{1/2} \ cis \left(\frac{\theta}{2} + \pi \right) \ . $$

The "tangent half-angle formula" gives us

$$ \tan \frac{\theta}{2} \ = \ \frac{\sin \theta}{1 + \cos \theta} \ = \ \frac{\frac{3}{5}}{1 + \frac{4}{5}} \ = \ \frac{1}{3} \ . $$

From this, we can derive $ \cos \frac{\theta}{2} = \frac{3}{\sqrt{10}} \ , \ \sin \frac{\theta}{2} = \frac{1}{\sqrt{10}} \ , \ $ and

$$ (4 + 3i)^{1/2} \ = \ \sqrt{5} \ \left( \ \frac{3}{\sqrt{10}} \ + \ i \ \frac{1}{\sqrt{10}} \right) \ , \ \sqrt{5} \ \left( \ -\frac{3}{\sqrt{10}} \ - \ i \ \frac{1}{\sqrt{10}} \right) $$

$$ = \ \ \frac{3 \sqrt{2}}{2} \ + \ i \ \frac{\sqrt{2}}{2} \ , \ -\frac{3 \sqrt{2}}{2} \ - \ i \ \frac{\sqrt{2}}{2} \ . $$

DeMoivre's Theorem produces the two complex square-roots of $ \ 4 + 3i \ , $ which we can now add to $ \ -i \ . $ This gives us the two roots of our original equation as

$$ z \ = \ \frac{3 \sqrt{2}}{2} \ + \ i \ \left( \frac{\sqrt{2}}{2} - 1 \right) \ \approx \ 2.121 \ - \ 0.293 \ i , $$

$$ = \ -\frac{3 \sqrt{2}}{2} \ - \ i \ \left( \frac{\sqrt{2}}{2} + 1 \right) \ \approx \ -2.121 \ - \ 1.707 \ i \ \ . $$

[The two zeroes are not complex conjugates of one another -- that "rule" only applies to zeroes of polynomials with real coefficients.]

Answer 2

$$x^2+2ix-(5+3i)=0\implies x_{1,2}=\frac{-2i\pm\sqrt{-4+4(5+3i)}}2$$

You now have to calculate the square root

$$\sqrt{-4+20+12i}=\sqrt{16+12i}=2\sqrt{4+3i}\;\ldots$$