Given Z1= sqr.root of 3+i and Z2= 6(cos(pie/6) + i sin(pie/6)) respond to the following:
*Convert Z1 to polar form
*Find Z1 dot Z2 (there is a dot in the middle) Leave your answer in polar form.
For those who understand complex numbers does not exists.
Now there are two square roots of !
But the radical sign is applied only to real numbers.
You seem to have some real confusion in mathematical notation and symbols.
What are you trying to ask?
BTW: Pie is a somthing one eats, like Amercian Pie. It is not a number.
Z1= sqr.root of 3+i and Z2= 6(cos(pie/6) + i sin(pie/6))
Okay, I don't know if you can take it from here, but Euler's Formula gives:
cos(x)+i sin(x)=e^(ix)
And:
sqrt.(3+i)=(3+i)^(1/2)=e^ln((3+i)^(1/2))=e^(.5(ln(3+i)))
And another formula gives:
ln(x+yi)=.5(ln(x^2+y^2))+i(arctan(y/x))
So Z(1)=sqrt.(3+i)=e^(.5(.5(ln(9+1))+i arctan(1/3))) (I may be missing a paranthesis, but you get the idea)
So now you have:
Z(1)=[e^(.25(ln(10)))][e^(i[.5(arctan(1/3)))]
Thus:
Z(1)=[e^(.25(ln(10)))][cos(.5(arctan(1/3)))+i sin(.5(artan(1/3)))]
And from there you can find the real and imaginary part of Z(1)
I may have made a typo somewhere in there, but the formulas are correct.