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I am teaching myself complex numbers, yet I can't solve this exercise:

Find all complex number solutions to $z^2+(1-i)z-1=0,$ and provide them in standard $z=a+bi$ format.

I've tried using the classic formula $\dfrac{-b \pm \sqrt{D} }{2a},$ and have found that $D=4-2i$, which leads to the solution: $$ z=-\frac{1}{2} +\frac{i}{2}\pm\sqrt{1-\frac{i}{2}}. $$

I am however unable to take the square root of either $1-\dfrac{i}{2}$ or $D=4-2i$, which means I cannot express the solution as $a+bi$.

I have tried inputting this into Wolfram Alpha, Symbolab and other online complex calculators and they all give different answers, none of which are the above answer, nor are they the same as the text gives, which is: $z=\dfrac{\pm\sqrt{\sqrt{5}+2}-1}{2}+\dfrac{1\pm\sqrt{\sqrt{5}-2}}{2}i.$

I do not see an obvious link between the given equation and the given solution. What am I missing?

amWhy
  • 209,954
Arnór Bjarki Svarfdal
  • 39

2 Answers2

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Your answer is correct. Note the principal square root of a complex number is

$$\sqrt{x+iy}=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+isgn(y)\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} .$$

Applied to your case gives $$\sqrt{1-i/2}=\frac{1}{2}\sqrt{2+\sqrt{5}}-\frac{i}{2}\sqrt{-2+\sqrt{5}} .$$

So the solution to your quadratic is

$$z=-\frac{1}{2} +\frac{i}{2}\pm\sqrt{1-\frac{i}{2}}\\ =\dfrac{-1\pm\sqrt{2+\sqrt{5}}}{2}+\dfrac{1\mp\sqrt{-2+\sqrt{5}}}{2}i$$

(note the ordering of the plus, minus signs).

Golden_Ratio
  • 12,591
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$$z^2+\left(1-i\right)z-1=0$$ $$\left(a+bi\right)^2+\left(1-i\right)\left(a+bi\right)-1=0$$ $$\left(a^2-b^2+2abi\right)+\left(a+bi-ai+b\right)=1$$ $$\left(a^2-b^2+a+b\right)+\left(2abi+bi-ai\right)=1$$ $$\left(a^2-b^2+a+b\right)+\left(2ab+b-a\right)i=1$$ $$Real\rightarrow a^2-b^2+a+b=1$$ $$Complex\rightarrow 2ab+b-a=0$$ Now, you just solve for $a$ and $b$ and it will be done. For study purposes, I will give the final answer. $$a=-1.529,b=0.743$$ $$a=0.529,b=0.257$$