Find the solutions of the equation $z^2+2z+1-i=0$ in the form$ z=p+iq$

Question

I had tried this so far : let $z = x + iy$ Hence, $$(x + iy)^2 + 2 (x + iy) + 1 - i = 0$$ $$x^2 + 2xyi + - y^2 + 2x + 2iy + 1 - i = 0$$ $$x^2 - y^2 + 2x + 1 + 2xyi + 2iy - i = 0$$

Answer 1

It's way easier to keep the complex variable whole

$$z^2+2z+1=(z+1)^2=i \implies z = -1\pm\sqrt{i} = -1\pm\frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}}$$

Answer 2

Gotta get used to the fact that complex numbers are actually numbers in their own right.

$z^2 +2z + 1-i=0$ will have solutions by the quadratic formula of

$z = \frac {-2 \pm \sqrt {4 - 4(1-i)}}2$

So factoring out the common factor $2$ we get $z = -1\pm i^{\frac 12}=$

$-1 \pm (\frac 1{\sqrt 2} + \frac 1{\sqrt 2}i) =$

$(-1\pm \frac 1{\sqrt 2}) + i(\pm \frac 1{\sqrt 2})$.

....

Although you could continue.

You must solve $x^2 - y^2 + 2x + 1=0$ and $2xy + 2y-1 = 0$.

So $y = x^2 +2x + 1$ so $y = (x+1)^2$ and $y = \pm(x+1)$.

But we also have $2xy + 2y - 1 = 0$ so $y(2x+2) = 1$ and $y = \frac 1{2(x+1)}$.

So we must solve $\pm (x+1) = \frac 1{2(x+1)}$ or

$\pm (x+1)^2 = \frac 12$ so $(x+1)^2 = \frac 12$ (and $y > 0$) and $x+1 =\pm \frac 1{\sqrt 2}$ and ... the rest falls out.

$x = -1 \pm \frac 1{\sqrt 2}$ and $y = \pm \frac 1{\sqrt 2}$.

(I actually wasn't expecting it to fall out so neatly.)

Answer 3

Along the line of what you were trying, it may be a bit easier (for this problem) to bring the "numerical" terms to one side of the equation, producing $$ z^2 \ + \ 2z \ \ = \ \ (p^2 + 2p - q^2) \ + \ (2pq + 2q)·i \ \ = \ \ -1 \ + \ i \ \ . $$ We can then set up the correspondence equations $$ p^2 + 2p - q^2 \ \ = \ \ -1 \ \ \ , \ \ \ 2·q·(p + 1) \ \ = \ \ 1 \ \ . $$ From the first of these, we obtain $ \ p^2 + 2p + 1 \ = \ ( p + 1 )^2 \ = \ q^2 \ \ . $ Squaring the second equation yields $ \ 4·q^2·(p + 1)^2 \ = \ 1 \ \ . $ Consequently, $$ \ 4·q^2·q^2 \ = \ 1 \ \Rightarrow \ \ q \ = \ \left(\frac14 \right)^{1/4} \ = \ \pm \frac{1}{\sqrt2} \ = \ \pm ( p + 1 ) \ \ . $$ It is evident, since the polynomial is quadratic, that two of the four possible combinations must be "spurious." Our second correspondence equation requires that $ \ (p + 1) \ $ and $ \ q \ $ must have the same sign, so the two zeroes of the polynomial are $ \ z \ = \ \left(-1 \pm \frac{1}{\sqrt2} \right) \ \pm \ \left(\frac{1}{\sqrt2} \right)·i \ \ . $