Answer:
Given the complex number: \(3(\cos 270^{\circ}+i \sin 270^{\circ})\)
Let \(z=3(\cos 270^{\circ}+i \sin 270^{\circ})\)
we have to write this complex number in the form of a+ib.
We know :
\(\cos 270^{\circ} = 0\)
\(\sin 270^{\circ} = -1\)
Then;
\(z = 3(0+i(-1)) = 3(0-i)\)
or
\(z = 0-3i\)
Therefore, the given complex number in the form of a+ib is, 0-3i
