Question

Write the complex number in the form a + bi.

square root of six(cos 315° + i sin 315°)

Answer 1

`√(3) -i√(3)`

Explanation:

The given complex number is,

`√(6)(\cos 315^(\circ) + i \sin 315^(\circ))`

we know that,

`\sin 315^(\circ)=-(1)/(\sqrt2)` and

`\cos 315^(\circ)=(1)/(\sqrt2)`

Putting the values,

`=√(6)\left((1)/(\sqrt2) -i(1)/(\sqrt2)\right)`

`=√(2)√(3)\left((1)/(\sqrt2) -i(1)/(\sqrt2)\right)`

`=√(2)√(3)\left((1)/(\sqrt2)\right) -i√(2)√(3)\left((1)/(\sqrt2)\right)`

`=√(3) -i√(3)`

Answer 2

`a=\sqrt6\cdot \cos 315=\sqrt 6 \cdot(\sqrt2)/(2)=(√(12))/(2)=(2\sqrt3)/(2)=\sqrt3\\ b=\sqrt6\cdot \sin 315=\sqrt 6 \cdot\left(-(\sqrt2)/(2)\right)=-(√(12))/(2)=-(2\sqrt3)/(2)=-\sqrt3\\\\ a+bi=\sqrt3-\sqrt3i`