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$$i^2=-1$$

$$a-b=1-i$$
$$a*b=2i-2$$

What might be "$b$"?

The answer is $2i$ can u explain how?

2 Answers 2

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Hint: $a =1-i+b$, hence $ab=2i-2 \implies (1-i+b)b=2i-2$. Solve this quadratic $$b^2+(1-i)b+(-2i+2)=0$$

by completing the square

$$b^2+2\frac{1-i}{2}b+(\frac{1-i}{2})^2-(\frac{1-i}{2})^2+(-2i+2)=0$$

or by the quadratic formula $$b_{1/2}= \frac{-(1-i)\pm \sqrt{(1-i)^2-4(-2i+2)}}{2}$$ to get values of $b$.

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    Unless you note and use the fact that $(1-i)^2-4(-2i+2)=6i-8=(1+3i)^2$, this is incomplete. – 2017-02-28
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    Thank you for this comment, but this can be derived by comparing $a+bi = (x+yi)^2=x^2-y^2+2xyi$ and then guessing / solving the associated quadratic. – 2017-02-28
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    ...Which is what I did to arrive at $(1+3i)^2$ and *which is missing from your post* (bis). Additionally, please be aware that the notation $\sqrt{z}$ should be avoided when $z$ is complex, not real positive. – 2017-02-28
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    I thought that this should be left to the OP. I know that $\sqrt{z}$ should be avoided, but for the quadratic formula, it doesn't make a difference because of the $\pm$. That is why I first said that completing the square should be used, but as I said it doesn't make a difference for a quadratic (as far as I know). – 2017-02-28
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    @Did it is not necessary to explain all steps, so long as the answer is clear enough, at least IMO. – 2017-02-28
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    @SimplyBeautifulArt Which steps are unnecessary depends on the level of the question. If one places oneself at a level such that the steps omitted here are obvious (which is the reason advanced to omit them, if I understand), it seems to me that the whole question becomes obvious. (A more minor point is that the notation $\sqrt{z}$ should be omitted.) – 2017-02-28
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$a=\frac{2i-2}{b}$

Put value of a in $a-b=1-i$

$\frac{2i-2}{b} - b = 1-i$

$(2i-2) - b^2 = (1-i)b$

$b^2 + (1-i)b - (2i-2) = 0$

Solve this quadratic equation to find b.