Question

Ignore the answer selected

Answer 1

Answer:

Option B is correct.

Step-by-step explanation:

$=6(cos270+isin270)\\\\=6(0-i)\\\\=6(-i)\\\\=-6i$

Answer 2

$\sf - 6i$

The modulus r, of a pure imaginary number bi, b<0, equals |b| and its argument θ equals $\frac{3\pi}{2}$

$\sf \: r = 6 \\ \sf \theta = \frac{3\pi}{2}$

Substitute the given values into the formula r(cos(θ)+i×sin(θ))

$\sf6( \cos( \frac{3\pi}{2} ) + i \times \sin( \frac{3\pi}{2} ) )$

Since, The value of 3π/2 is 270°.. Put it in the equation

$\boxed{ \tt6( \cos270 + i \times \sin270) }$

Thus, Option B is the correct choice!!~