# Complex Numbers

• Mar 5th 2010, 09:43 PM
demode
Complex Numbers
Put the following expressions into the shape $a+ib$ with $a,b \in \mathbb{R}$.

(a) $i^3$

(b) $(1-i)i$

(a) $-i$

(b) $1+i$

I know that the real and imaginary parts of a+ib are a and b respectively. But I don't know how to arrive at this answers, can anyone please explain? Thanks.
• Mar 5th 2010, 09:50 PM
TheEmptySet
Quote:

Originally Posted by demode
Put the following expressions into the shape $a+ib$ with $a,b \in \mathbb{R}$.

(a) $i^3$

(b) $(1-i)i$

(a) $-i$

(b) $1+i$

I know that the real and imaginary parts of a+ib are a and b respectively. But I don't know how to arrive at this answers, can anyone please explain? Thanks.

Remember the defintion of i is $i^2=-1$

For the first $i^3=i^2\cdot i$

For the 2nd just distribute and use the property above.
• Mar 5th 2010, 11:21 PM
demode
Quote:

Originally Posted by TheEmptySet
Remember the defintion of i is $i^2=-1$

For the first $i^3=i^2\cdot i$

For the 2nd just distribute and use the property above.

Thank you, that helps a lot. But I don't understand why $i^2=-1$! When I look at the imaginary-real axis, $i$ on the imaginary axis corresponds to 1 on the real axis. That's why I thought $i^3 = 1^3$.
• Mar 6th 2010, 02:11 AM
Prove It
Quote:

Originally Posted by demode
Thank you, that helps a lot. But I don't understand why $i^2=-1$! When I look at the imaginary-real axis, $i$ on the imaginary axis corresponds to 1 on the real axis. That's why I thought $i^3 = 1^3$.

$i^2 = -1$ by DEFINITION of imaginary and complex numbers.

This is because they needed to create a number (i.e. out of their imaginations) that $= \sqrt{-1}$. They called this number $i$.

So if $i = \sqrt{-1}$ then that must mean $i^2 = -1$.