1. ## complex numbers

i am stuck on question 12 I worked out part (ii) by saying z^3 = x(cos360 +isin360) therefore z=x(cos120+isin120). i am not sure how to do part (iii)

2. ## Re: complex numbers

Originally Posted by edwardkiely

i am stuck on question 12 I worked out part (ii) by saying z^3 = x(cos360 +isin360) therefore z=x(cos120+isin120).
This is incorrect. It should be $z= x^{1/3}W(cos(120)+ i sin(120))$.

If $z= x(cos(\theta)+ i sin(\theta))$ then $|z|= x$ so [tex]|^3|= (9/4)|z| then [tex]x^3= (9/4)x. 4x^3- 9x= x(4x^2- 9)= 0. Can you solve for x?

i am not sure how to do part (iii)

3. ## Re: complex numbers

what does W stand for?

i am sorry but i can not understand you last line.

4. ## Re: complex numbers

Originally Posted by edwardkiely
what does W stand for?
i am sorry but i can not understand you last line.
That W is just a miss-type of a random symbol.

Here is what is going on:
\begin{align*}\text{Because }|z^3|&=|z|^3 \\|z|^3&=\dfrac{9}{4}|z|\\|z|^2&=\dfrac{9}{4}\\|z| &=\dfrac{\pm 3}{2} \end{align*}

Can you now finish?

5. ## Re: complex numbers

Originally Posted by HallsofIvy
This is incorrect. It should be $z= x^{1/3}W(cos(120)+ i sin(120))$.

If $z= x(cos(\theta)+ i sin(\theta))$ then $|z|= x$ so [tex]|^3|= (9/4)|z| then [tex]x^3= (9/4)x. 4x^3- 9x= x(4x^2- 9)= 0. Can you solve for x?
Thanks, Plato. Yes, the "W" was a mistype. It should have been $z= x^{1/3}(cos(120)+ i sin(120))$. The point is that, with |z|= x, you should have had " $x^{1/3}$" not " $x$".

$x^3= (9/4)x$ is equivalent to $4x^3- 9x= x(4x^2- 9)= 0$ so either $x= 0$ or $4x^2- 9= (2x- 3)(2x+ 3)= 0$.

Of course, in order to have |z|= x, with z non-zero as shown, x must be a positive real number.

thanks