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