# Thread: complex , is this correct?

1. ## complex , is this correct?

(1 + i)^3(sqrt(3) + i)^3= (sqrt(8))^3* (cos 15pi/12 + i sin 15pi/12)

and (1 + i)^3(sqrt(3) + i)^3= -16-i16

2. Originally Posted by sandy
(1 + i)^3(sqrt(3) + i)^3= (sqrt(8))^3* (cos 15pi/12 + i sin 15pi/12)

and (1 + i)^3(sqrt(3) + i)^3= -16-i16
Did you convert the left hand sides into polar forms and then apply deMoivres Theorem and the usual rule of multiplying complex numbers written in polar form?

It's more efficient to check answers if you show all steps of your working.

3. Originally Posted by mr fantastic
Did you convert the left hand sides into polar forms and then apply deMoivres Theorem and the usual rule of multiplying complex numbers written in polar form?

It's more efficient to check answers if you show all steps of your working.
No i multiplied them first then to the power of 3 then i used deMoivres Theorem.
It's wrong isnt it oooooooooooohhhhhhhhhhh
what should i do

4. Originally Posted by sandy
No i multiplied them first then to the power of 3 then i used deMoivres Theorem.
It's wrong isnt it oooooooooooohhhhhhhhhhh
what should i do
Do what I said to do in my first post.

Then show all your working.

5. Originally Posted by mr fantastic
Do what I said to do in my first post.

Then show all your working.
here is what i done

(1+i)^3=2*sqrt(2)( cos 3pi/4+ i sin 3pi/4)

(sqrt3 + i)^3=8 ( cos 3pi/6 +i sin 3pi/6)

so multiplying these two we get r1*r2= 16*sqrt(2)

and cos (theta1 + theta2)=cos (15pi/12)
and sin (theta1 + theta2)=sin (15pi/12)

so (1+i)^3(sqrt3 + i)^3=16*sqrt(2)(cos (15pi/12) +i sin (15pi/12))
= -16 - i 16

is that right?

6. Originally Posted by sandy
here is what i done

(1+i)^3=2*sqrt(2)( cos 3pi/4+ i sin 3pi/4)

(sqrt3 + i)^3=8 ( cos 3pi/6 +i sin 3pi/6)

so multiplying these two we get r1*r2= 16*sqrt(2)

and cos (theta1 + theta2)=cos (15pi/12)
and sin (theta1 + theta2)=sin (15pi/12)

so (1+i)^3(sqrt3 + i)^3=16*sqrt(2)(cos (15pi/12) +i sin (15pi/12))
= -16 - i 16

is that right?
Yes. (Although I would have simplified 3pi/6 to pi/2. And note that 15pi/12 is not the principle argument, although it doesn't matter here).