Solve this complex number: $z^2+(1+i)z+i=0$

Question

I need to solve:

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

first of all I used $$z = (a+ib)$$ and I get:

$$(a+ib)^2+(1+i)(a+ib)+i=0$$ $$a^2-b^2+2aib+a+ib+ia-b+i=0$$

then I have ordered, on the left the number without the $i$ and on the right all the number with the $i$:

$$\underbrace{a^2-b^2+a-b}_\text{real} + \underbrace{2aib+ib+ia+i}_\text{imaginary} = 0$$

So I have set this:

$$\begin{cases} a^2-b^2+a-b=0 \\ 2aib+ib+ia+i \end{cases}$$

$$\begin{cases} a(a+1)-b(b-1) = 0 \\ 2aib+ib+ia+i \end{cases}$$

How can I procede? I can say that the first equation is true when $a=b$

Hint We can factor the original quadratic readily, and this amounts to solving: What are two numbers whose sum is $1 + i$ and whose product is $i$? — Travis Willse, Jan 18 '16 at 10:14
Why not just use the quadratic formula? It's valid for complex numbers too! — najayaz, Jan 18 '16 at 10:15
Of course @Travias. Vietta's Theorem! Indeed it is easier than quadratic formula — sinbadh, Jan 18 '16 at 10:15
@PM2Ring: I think Travis was looking for factors like $(z+\cdots)(z+\cdots)$ rather than roots — Henry, Jan 18 '16 at 10:29
@PM2Ring What Henry said, but of course the problems are equivalent. — Travis Willse, Jan 18 '16 at 10:30
You might have a look at similar older posts, such as http://math.stackexchange.com/questions/1487813/how-to-solve-this-z2-1-3iz-2i-2-0 and http://math.stackexchange.com/questions/1419221/find-solution-to-complex-equation — Martin Sleziak, Jan 18 '16 at 11:33

score 5 · Accepted Answer · answered Jan 18 '16 at 10:17

5

hint: $z^2 + (1+i)z + i = 0 \Rightarrow z^2+z + iz+i = 0\Rightarrow z(z+1) + i(z+1) = 0 \Rightarrow ...$

answered Jan 18 '16 at 10:17

DeepSea

77,651

orion · Answer 2 · 2016-01-18T13:52:47.107

The others pointed out a few ways how to guess the solution or derive the result for this specific case. However, the beauty of complex numbers is also, that math works on them just fine. So... just use the quadratic formula

$$z=\frac{-(1+i)\pm \sqrt{(1+i)^2-4i}}{2}=\frac{-(1+i)\pm \sqrt{-2i}}{2}$$ $$=\frac{-(1+i)\pm (1-i)}{2}=\{-i,-1\}$$ I used $\sqrt{-2i}=\sqrt{2e^{-i\pi/2}}=\sqrt{2}(\cos(-\pi/4),\sin(-\pi/4))=1-i$.

score 0 · Answer 3 · answered Jan 18 '16 at 10:21

0

Cool thing about complex polynomials is that they alwas have n solutions (where n is the order of the polynomial) and you can brute force it quite simply. $$z^2 + (1+i)z + i = 0 \implies$$ $$(z+\frac{1+i}{2})^2 = -i - (1+i)^2 = -3i = 3e^{\frac{-\pi}{2}i}$$

Now you can take a square root to get:

$$z +\frac{1+i}{2} = \sqrt{3}e^{\frac{-\pi}{2}i \pm \pi i}$$

Simplify and solve.

answered Jan 18 '16 at 10:21

Piotr Benedysiuk

1,622

I think you should have $(z+\frac{1+i}{2})^2 = -i + \left(\frac{1+i}{2}\right)^2 = \left(\frac{1-i}{2}\right)^2$. Your completing the square is equivalent to the quadratic formula – Henry Jan 18 '16 at 10:26
You are right. my bad. – Piotr Benedysiuk Jan 18 '16 at 10:52

Solve this complex number: $z^2+(1+i)z+i=0$

3 Answers3