Question

Write the complex numbers `4cis120^@` and `9cis((3pi)/2)` in the form `a+ib`?

Answer 1

See explanation.

Explanation:

I assume that the "cis" is an abbreviation of `(cos + i sin)`. If we use the formula we get:

`Bi`

`4*(cos120+isin120)=4*(cos(90+30)+isin(90+30))`

`=4*(-sin30+icos30)=4*(-1/2+i*sqrt3/2)=-2+2sqrt3i`

As real component is negative and imaginary component is positive, this lies in second quadrant.

`Bii`

`9*(cos((3pi)/2)+isin((3pi)/2))=9*(0+i*(-1))=-9i`

As real component is zero and imaginary component is negative, this lies on imaginary axis, `9` points below `0+i0`.

The sketched numbers are as shown below:

graph{(x^2+(y+9)^2-0.12)((x+2)^2+(y-2sqrt3)^2-0.1)=0 [-20.5, 19.5, -12.44, 7.56]}