# Math Help - Complex numbers

1. ## Complex numbers

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.

2. For those who understand complex numbers $\sqrt {3 + i}$ does not exists.
Now there are two square roots of ${3 + i}$!
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.

3. i just wrote everything just like my teacher gave it to me, and i wrote pie because i don't know how to put the symbol.

4. Originally Posted by Cyberman86
i just wrote everything just like my teacher gave it to me, and i wrote pie because i don't know how to put the symbol.
The Greek letter is pi, $\pi$.
It is realy sad but your ‘teacher’ may not be up to speed.
Those easily misunderstand complex numbers or Complex Analysis who are minimally prepared.

5. ## Help

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.