Question

Find z1z2 if z1 = 3(cos37° + isin37°) and z2 = 2/3(cos53° + isin53°).

Answer 1

`z_(1)\cdot z_(2) = 2\cdot (\cos 90^(\circ) + i \cdot \sin 90^(\circ))`

Explanation:

Both variable can be rewritten into polar form:

`z_(1) = 3\cdot e^(i\cdot 0.205\pi)` and
`z_(2) = (2)/(3)\cdot e^(i\cdot 0.294\pi)`

The complex product is equal to:

`z_(1)\cdot z_(2) = (3)\cdot \left((2)/(3)) \cdot e^(i\cdot (0.205\pi+0.294\pi))`

`z_(1)\cdot z_(2) = 2 \cdot e^(i\cdot 0.499\pi)}`

The resultant expression in rectangular form is:

`z_(1)\cdot z_(2) = 2\cdot (\cos 90^(\circ) + i \cdot \sin 90^(\circ))`

Answer 2

The correct question is:

Find
`z_(1).z_(2)` if
`z_(1)= 3(cos37 + isin37)` and
`z_(2)= (2)/(3) (cos53 + isin53)`.

(Note: the angles mentioned in the equations above are in degrees)

The answer is
`z_(1).z_(2)= 2i`

Explanation:

`z_(1).z_(2)= 3(cos37 + isin37)* (2)/(3) (cos53 + isin53)`

`z_(1).z_(2)= (3cos37 + 3isin37)* ((2)/(3) cos53 + (2)/(3)isin53)`

`z_(1).z_(2)= 3cos37*(2)/(3) cos53 + 3cos37*(2)/(3)isin53+3isin37*(2)/(3) cos53+3isin37*(2)/(3)isin53`

`z_(1).z_(2)= 0.961+1.275i+0.724i-0.961` (Because
`i*i=i^(2)=-1` &
`i=√(-1)`)

`z_(1).z_(2)= 0.961+1.275i+0.724i-0.961`

`z_(1).z_(2)= 2i`