What's the result? $1/i=?$, where $i=\sqrt{-1}$ 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!
 A: 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 $a,b>0.$
A: 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
A: 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?
A: Hint: $\frac{1}{i}=\frac{1}{i}\cdot\frac{-i}{-i}$
A: $\displaystyle x=\frac{1}{i}$ is such a number, that $x\cdot i=1$. Note that $i \cdot (-i)=(-1) \cdot i^2=(-1) \cdot (-1)=1$, so $\frac{1}{i}=-i$.
A: 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.
