How do you express the complex number in trigonometric form: #-1+i#? Precalculus Complex Numbers in Trigonometric Form Trigonometric Form of Complex Numbers 1 Answer Konstantinos Michailidis May 27, 2016 We have that #-1+i=sqrt2*(-sqrt2/2+sqrt2/2*i)=sqrt2*(cos((3pi)/4)+i*sin((3pi)/4))# Answer link Related questions How do I find the trigonometric form of the complex number #-1-isqrt3#? How do I find the trigonometric form of the complex number #3i#? How do I find the trigonometric form of the complex number #3-3sqrt3 i#? How do I find the trigonometric form of the complex number #sqrt3 -i#? How do I find the trigonometric form of the complex number #3-4i#? How do I convert the polar coordinates #3(cos 210^circ +i\ sin 210^circ)# into rectangular form? What is the modulus of the complex number #z=3+3i#? What is DeMoivre's theorem? How do you find a trigonometric form of a complex number? Why do you need to find the trigonometric form of a complex number? See all questions in Trigonometric Form of Complex Numbers Impact of this question 1962 views around the world You can reuse this answer Creative Commons License