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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 $xi +y$?

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    $\begingroup$ Is it not just $1$? $\endgroup$ Dec 1, 2016 at 1:02
  • $\begingroup$ @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 $$\frac{i}{i} = 1 \ \text{because numerator and denominator are the same}$$? $\endgroup$
    – BCLC
    Dec 1, 2016 at 6:56

2 Answers 2

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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.

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  • $\begingroup$ 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 $$\frac{i}{i} = 1 \ \text{because numerator and denominator are the same}$$? $\endgroup$
    – BCLC
    Dec 1, 2016 at 6:56
  • $\begingroup$ 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. $\endgroup$
    – Oiler
    Dec 1, 2016 at 13:57
  • $\begingroup$ Oiler, I'm not quite sure I understand you. You know I'm agreeing with you right? $\endgroup$
    – BCLC
    Dec 2, 2016 at 5:30
  • $\begingroup$ Okay then I do not understand your original comment. What are you asking? $\endgroup$
    – Oiler
    Dec 2, 2016 at 16:03
  • $\begingroup$ we should follow your way of multiplying by conjugate instead of just saying = 1 because numerator and denominator are the same? $\endgroup$
    – BCLC
    Dec 2, 2016 at 19:58
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$$ \frac{1}{i} \cdot i = \frac{i}{i} = 1 = 1+0i$$

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    $\begingroup$ Why are you using $\iff$ instead of $=$? $\endgroup$ Dec 1, 2016 at 1:10
  • $\begingroup$ @Omnomnomnom Does it matter in this case? Just wondering haha $\endgroup$ Dec 1, 2016 at 1:12
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    $\begingroup$ $\iff$ is used for statements, things that are true or false $\endgroup$
    – GFauxPas
    Dec 1, 2016 at 1:13
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    $\begingroup$ @GFauxPas Oh sorry, my bad! Will change it :) $\endgroup$ Dec 1, 2016 at 1:13
  • $\begingroup$ 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 $$\frac{i}{i} = 1 \ \text{because numerator and denominator are the same}$$? $\endgroup$
    – BCLC
    Dec 1, 2016 at 6:56

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