# Could someone please help me write $z= \frac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form?

Could someone please help me write $z = \dfrac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form?

$|z|=\sqrt{2}$

But how do I proceed to get the argument?

$\mathrm{arg⁡}(z)=\mathrm{arg⁡}(2+2i)^3 - \mathrm{arg⁡}(1+i\sqrt{3})^4$

Thanks /David

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Welcome to math.stackexchange! Why don't you break this problem into smaller pieces first: Find the polar forms of $2+2i$ and $1+i\sqrt{3}$ first. Then can you find the full answer? – Ragib Zaman Nov 4 '11 at 15:45
For integer $n,\arg(z^n)=n\arg(z)$, which you can see from writing $z=r\exp(i\theta)$ – Ross Millikan Nov 4 '11 at 15:47
@David: Another hint: Let $z_1= |z_1|exp(i \alpha)$ and $z_2= |z_2|exp(i \beta)$. Then $z_1/z_2= (|z_1|/|z_2|)exp (i (\alpha - \beta))$. – HbCwiRoJDp Nov 4 '11 at 16:29

Hints: Use $\arctan 1=\frac{1}{4}\pi$ and $\arctan \sqrt{3}=\frac{1}{3}\pi$ in

$$\arg \left( \left( 2+2i\right) ^{3}\right) =3\arg \left( 2+2i\right) =3 \arctan \frac{2}{2}=3\times \frac{1}{4}\pi$$

and

$$\arg \left( (1+i\sqrt{3})^{4}\right) =4\arg (1+i\sqrt{3})=4\arctan \frac{\sqrt{3}}{1}=4\times \frac{1}{3}\pi .$$

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Hint: The argument of $1+i$ is $\frac{\pi}{4}$ by looking at the angle of the special triangle whose sides are $1$, $1$ and hypotenuse $\sqrt{2}$. Also the argument of $1+i\sqrt{3}$ is $\frac{\pi}{3}$ by looking at the angle of the special triangle whose sides are $1$, $\sqrt{3}$ and hypotenuse $2$ . Can you solve it from here?

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You could just calculate and simplify the fraction:

$$(2+2i)^3=8(1+i)^3=8\cdot (2i) \cdot (1+i)=-16+16i$$

and

$$(1+i\sqrt{3})^4=(-2+2\sqrt{3}i)^2=-8-8\sqrt{3}i$$

Amplify with the conjugate of the denominator and find the algebraic form of the complex number. Sometimes, when you don't remember the tricks presented in the other answers, it's easier to do the computation directly and only then transform into trigonometric form.

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