*****25 POINTS*****Write the complex number in the form a + bi. square root of six(cos 315° + i sin 315°)A) square root of three over two minus square root of three over two times iB) square root of six minus square root of sixiC) square root of three minus square root of threeiD) square root of six over two minus square root of six over two times i

Question
Answer:
The initial expression is

√6 *(cos 315° + i sin 315°)

We just multiply the factor to express it in the a + ib form

√6 *cos 315° + i √6 *sin 315°

This is equal to √6 *(√2/2) +i √6 *(-√2/2) = √3 -i√3

solved
general 5 months ago 4604