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This was a test problem that I did not understand at all. I know it is converting complex numbers, but I need help.

How do I write $\frac{1}{i}i$ in the form $xi +y$?

Tsangares
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    Is it not just $1$? – ImHereSometimes Dec 01 '16 at 01:02
  • @ImHereSomtimes Cmiiw, but i/i needs to be evaluated as follows $$\frac{i}{i} \frac{-i}{-i} = \frac{-i^2}{-i^2} = \frac{1}{1} = 1$$ and not as follows $$\frac{i}{i} = 1 \ \text{because numerator and denominator are the same}$$? – BCLC Dec 01 '16 at 06:56

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In general, if you have $$ \frac{a+bi}{c+di} $$ you can multiply by $ (c-di)/(c-di)$ and simplify things nicely... I'll leave the details to you, but can you see how to use this for your problem? But as people have pointed out, it is indeed just equal to 1.

Oiler
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  • Yours is correct though since i/i needs to be evaluated as follows $$\frac{i}{i} \frac{-i}{-i} = \frac{-i^2}{-i^2} = \frac{1}{1} = 1$$ and not as follows $$\frac{i}{i} = 1 \ \text{because numerator and denominator are the same}$$? – BCLC Dec 01 '16 at 06:56
  • While $\frac{z}{z} = 1$ for any $z \in \mathbb{C}$ that is nonzero, the method I proposed is a more general method for dividing complex numbers. In fact, you could derive $\frac{z}{z} = 1$ by using this method. Truthfully, it can be evaluated either way. – Oiler Dec 01 '16 at 13:57
  • Oiler, I'm not quite sure I understand you. You know I'm agreeing with you right? – BCLC Dec 02 '16 at 05:30
  • Okay then I do not understand your original comment. What are you asking? – Oiler Dec 02 '16 at 16:03
  • we should follow your way of multiplying by conjugate instead of just saying = 1 because numerator and denominator are the same? – BCLC Dec 02 '16 at 19:58
  • Yes. This is the first principles way to divide complex numbers and will work regardless of the numbers involved. – Oiler Dec 02 '16 at 20:01
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$$ \frac{1}{i} \cdot i = \frac{i}{i} = 1 = 1+0i$$

bigfocalchord
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