(-1+i)/(1+i) What is this in standard notation?

Question

842 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T04:23:13

#i#

.

This a division of one complex (imaginary) number by another. In complex numbers, you need to know that the standard form of a complex number is:

#a+-bi#

#i=sqrt(-1)#

#i^2=-1#

We need to multiply the top and the bottom by the conjugate of the bottom:

#((-1+i))/((1+i))=((-1+i)(1-i))/((1+i)(1-i))=(-1+i+i+1)/(1+1)=(2i)/2=i#

In the form #a+-bi#, #a# is the real part and #b# is the imaginary part.

In our problem:

#a=0, b=+1#

Our number is:

#0+1i# which is #i#