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I had to solve: $z^2 + 1 - i = 0$. What I did:

$$z_{1, 2} \frac{-0 \pm \sqrt{0 - 4(1 - i)}}{2} = \pm \sqrt{i - 1}$$

Now, I need to find $\Im(z_{1, 2})$ and $\Re(z_{1,2})$.

I think there is a way to solve it using the trigonometric representation, right? I somehow didn't found what I wanted so I'd love if someone can just tell me what I should look up. I'm okay with you people just giving me the proper term to google. :)

Thanks

Michael Rozenberg
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xotix
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4 Answers4

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use the Ansatz $$\sqrt{-1+i}=a+bi$$

Dr. Sonnhard Graubner
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Let $-1+i=(x+yi)^2$, where $x$ and $y$ be real numbers.

Thus, $$x^2-y^2=-1$$ and $$2xy=1.$$ Thus, $y=\frac{1}{2x}$ and $$x^2-\frac{1}{4x^2}=-1$$ or $$4x^4+4x^2-1=0$$ or $$x^2=\frac{-2+\sqrt8}{4},$$ which gives $x=\sqrt{\frac{\sqrt2-1}{2}}$ or $x=-\sqrt{\frac{\sqrt2-1}{2}}$ and we got the answer: $$\sqrt{-1+i}=\left\{\sqrt{\frac{\sqrt2-1}{2}}+\sqrt{\frac{\sqrt2+1}{2}}i, -\sqrt{\frac{\sqrt2-1}{2}}-\sqrt{\frac{\sqrt2+1}{2}}i\right\}.$$ Also, we have $$-1+i=\sqrt2\left(-\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i\right)=\sqrt2\left(\cos135^{\circ}+i\sin135^{\circ}\right).$$ Id est, $$\sqrt{-1+i}=\left\{\sqrt[4]2\left(\cos67.5^{\circ}+i\sin67.5^{\circ}\right),-\sqrt[4]2\left(\cos67.5^{\circ}+i\sin67.5^{\circ}\right)\right\},$$ which is the same answer of course.

Michael Rozenberg
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I'd avoid writing the square root $\sqrt{}$ function when you deal with exponentials, because $\sqrt{x}$ is only defined for nonnegative real numbers (because for example, it is unclear whether $\sqrt{-1}=i$ or $\sqrt{-1}=-i$).

Instead, you should be dealing with the equation $$z^2=i-1$$ directly.

You can find the solutions to that equation by writing $z=r\cdot e^{i\phi}$ where $r\geq 0$ and $\phi\in[0,2\pi)$.

The equation then becomes $$r^2e^{i(2\phi)} = i-1$$

Now, you can also write $i-1$ in polar coordinates, as $i-1 = r_s \cdot e^{i\phi_s}$ (I'll leave the calculation of $r_s$ and $\phi_s$ to you), and you are left with the equation

$$r^2e^{i(2\phi)} = r_s \cdot e^{i\phi_s}$$ from which you can get $r=\sqrt{r_s}$ and the two possible values of $\phi$.

5xum
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  • From the OP's comments it looks like it might be useful for him to be told explicitly that $e^{i\phi} = \cos(\phi)+i \sin(\phi)$. – hmakholm left over Monica Oct 03 '17 at 13:33
  • you actually pointed me to another little problem. I also solved $z^3=i$ for which I guessed the solution $z_1 = -\sqrt{-1}$. Hmm, I'm gonna look at it again an try to resolve it. Also try your appraoch for the square root calculation. Thanks – xotix Oct 03 '17 at 13:33
  • @xotix There are three solutions to the equation $z^3=i$. – 5xum Oct 03 '17 at 13:34
  • @HenningMakholm yeah I know that. But thanks! :) I do have 3 solutions. I guessed one for getting the other twos. – xotix Oct 03 '17 at 13:34
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I'll give a simpler way of computing the square roots of complex numbers in algebraic form: $$(x+iy)^2=-1+i\iff\begin{cases}x^2-y^2=-1\\2xy=1\end{cases}$$ Furthermore, $$|x+iy|^2=|-1+i|\iff x^2+y^2=\sqrt2.$$ This will reduce the problem to a simple linear system in $x^2$, $y^2$: $$\begin{cases}x^2+y^2=\sqrt2\\x^2-y^2=-1\end{cases}\iff\begin{cases}x^2=\dfrac{\sqrt 2-1}2\\[1ex]y^2=\dfrac{\sqrt 2+1}2\end{cases}$$ Note that $\;xy=\dfrac{1}{2}$ implies $x,y$ have the same sign, so $$x+iy=\pm\frac{\sqrt2}2\biggl(\sqrt{\sqrt{\mathstrut2\,}-1}+i\sqrt{\sqrt{\mathstrut2\,}+1}\biggr).$$

Bernard
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