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I just had my first math class in the university, and I understood everything pretty well, but I think I have misread this one because I read that the result is $-1$. Thanks for your answers!

user153012
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majjjom
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6 Answers6

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Consider: $$\dfrac{1}{i}=\dfrac{1}{i}\times\dfrac{i}{i}=\dfrac{i}{-1}=-i$$ However, the following argument would not work: $$\dfrac{1}{i}=\dfrac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\dfrac{1}{-1}}=\sqrt{-1}=\pm i .$$

The latter argument fails because $\frac{\sqrt{a}}{\sqrt{b}} \equiv \sqrt{\frac{a}{b}}$ holds (if and) only if $a,b>0.$

beep-boop
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  • Dear @alexqwx, than you for the edit. Meanwhile, I would like to ask the downvoter why the answer was downvoted, so that I could improve the answer accordingly. –  Sep 08 '14 at 22:21
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    Sure! The downvote wasn't from me btw; I think it's a great answer. I just added that last sentence as the cherry on the cake. – beep-boop Sep 08 '14 at 22:35
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by definition: $i\cdot i =-1$ by dividing by $i$, as in basic algebra: $i=-\frac{1}{i}$ and then $\frac{1}{i} =-i$

Cheers

UserX
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henry G
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Hint:

$$\frac{1}{i} = 1\cdot \frac{1}{i} = \frac{i}{i}\cdot \frac1i = \frac{i\cdot 1}{i\cdot i}$$ can you take it from here?

5xum
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Hint: $\frac{1}{i}=\frac{1}{i}\cdot\frac{-i}{-i}$

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$\displaystyle x=\frac{1}{i}$ is such a number, that $x\cdot i=1$. Note that $i \cdot (-i)=(-1) \cdot i^2=(-1) \cdot (-1)=1$, so $\frac{1}{i}=-i$.

agha
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Have they taught you the rule for dividing complex numbers yet? $$\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$$ (see http://mathworld.wolfram.com/ComplexDivision.html).

Of course $1$ and $i$ are not complex numbers, but you can use the rule easily enough by adding in zeroes: $$\frac{1 + 0i}{0 + 1i} = \frac{(0 + 0) + (0 - 1)i}{0^2 + 1^2} = \frac{-i}{1} = -i.$$

Consider also the following:

  • Multiplication by $1$ leaves the real and imaginary parts exactly the same, in value and in sign.
  • Multiplication by $i$, causes the real and imaginary parts to trade places, and the sign of the new real part is opposite the sign of the old imaginary part.
  • Multiplication by $−1$ toggles the sign of the real part and toggles the sign of the imaginary part.
  • Multiplication by $-i$, causes the real and imaginary parts to trade places, and the sign of the new imaginary part is opposite the sign of the old real part.

(from http://oeis.org/wiki/Complex_numbers#Complex_units_and_identity_elements).

So then $-i \times i = -(-1) = 1$ (the real and imaginary parts of the multiplicand trade places, and the sign of the real part of the result is opposite the sign of the imaginary part of the multiplicand), confirming the result $\frac{1}{i} = -i$ above.

James47
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    Nice answer!

    $1$ and $i$ are considered complex numbers, sorry. You have your terminology a bit confused. (The phrase "complex numbers" includes the reals and the imaginaries, in addition to everything else complex. Otherwise, we couldn't say "the sum/product of any two complex numbers is a complex number.")

     – Akiva Weinberger Sep 08 '14 at 23:15
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    That little gaffe aside, I think this deserves an upvote. Maybe he could rewrite that one sentence to say something like "Even though 1 and $i$ have no imaginary or real parts, respectively, you can still use the rule for complex division by putting in a few zeroes..." – Robert Soupe Sep 09 '14 at 01:15