How do you write the complex number in trigonometric form #3-i#?

Question

3261 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T16:12:56

The answer is #=sqrt10(cos(-18.4º)+isin(-18.4º))#

The trigonometric form of a complex number

#z=a+ib#

is

#z=r(cos theta + i sin theta)#

#rcostheta=a#

#rsintheta=b#

#r^2=a^2+b^2=|z|^2#

Here, we have

#z=3-i#

#r=|z|=sqrt(9+1)=sqrt10#

#z=sqrt10(3/sqrt10-i/sqrt10)#

#costheta=3/sqrt10#

#sin theta=-1/sqrt10#

So, we are in the fourth quadrant

#theta=-18.4#º

#z=sqrt10(cos(-18.4º)+isin(-18.4º))#