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

Question

3431 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

$z=sqrt10(cos(-18.4º)+isin(-18.4º))$

How do you write the complex number in trigonometric form 3-i3-i?