Dividing Complex Numbers

[masiigymnast]

Hello, I'm new here, so I'm not really sure if this question has been asked before and whether it is allowed to double-post past questions, so sorry if this is a repeat post.

One of the questions in my textbook in the complex number section is "simplify 1/i "

I figured to convert 1 into 1+0i, and i into 0 +1i, and then follow the standard way of dividing complex numbers (a+bi)/(c+di)= (ac-bd)+ (ad-bc)i, and I keep ending up with i/1, although I'm supposed to get -i. Can someone please share how to go about this? Thank you so much!

Masha

 

[HallsofIvy]

Hello, I'm new here, so I'm not really sure if this question has been asked before and whether it is allowed to double-post past questions, so sorry if this is a repeat post.

One of the questions in my textbook in the complex number section is "simplify 1/i "

I figured to convert 1 into 1+0i, and i into 0 +1i, and then follow the standard way of dividing complex numbers (a+bi)/(c+di)= (ac-bd)+ (ad-bc)i, and I keep ending up with i/1, although I'm supposed to get -i. Can someone please share how to go about this? Thank you so much!

Masha

The real question is "how in the world did you get i/1?" In any case, "(a+bi)/(c+di)= (ac-bd)+ (ad-bc)i" is certainly wrong!

You divide complex numbers by multiplying both numerator and denominator by the complex conjugate of the denominator:
$\displaystyle \frac{a+ bi}{c+ di}= \frac{a+ bi}{c+ di}\frac{c- di}{c- di}= \frac{(ac- bd)+ (bc- ad)i}{c^2+ d^2}$

Your formula has the wrong sign on the imaginary part and, most importantly, is missing the denominator entirely!

Here, with $\displaystyle \frac{1}{i}$, the denominator is i and its complex conjugate is -i. So multiply both numerator and denominator by -i:
$\displaystyle \frac{1}{i}\frac{-i}{-i}$. What does that give you?