How do you write the trigonometric form of #-3-i#?

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Answer 1 · 2017-08-01T13:44:06

The trigonometric form is #=sqrt10(cos198.4^@+isin198.4^@)#

Let #z=-3-i#

The trigonometris form is

#z=r(costheta+isintheta)#

If #z=a+ib#

#z=|z|(a/|z|+b/|z|i)#

The modulus is

#|z|=sqrt(a^2+b^2)=sqrt((-3)^2+(-1)^2)=sqrt10#

Therefore,

#z=sqrt10(-3/sqrt10-1/sqrt10i)#

#costheta=-3/sqrt10#

and

#sintheta=-1/sqrt10#

We are in the quadrant #III#

#theta=198.4^@#

Therefore,

#z=sqrt10(cos198.4^@+isin198.4^@)=e^(i198.4^@)#