How do you express the complex number in trigonometric form: #6#? Precalculus Complex Numbers in Trigonometric Form Trigonometric Form of Complex Numbers 1 Answer Daniel L. Jun 8, 2018 See explanation. Explanation: The number #6# lies on a positive part of real axis, so the angle is #0# and the module is #6#, so the trigonometric form of #6# is: #6=6*(cos0^o + isin0^o)# ## 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 7656 views around the world You can reuse this answer Creative Commons License