# Homework Help: Help with finding the modulus, polar form and polar exponential form

1. Apr 24, 2012

### Cottontails

1. The problem statement, all variables and given/known data
Consider the complex number z=(i^201+i^8)/(i^3(1+i)^2).
(a) Show that z can be expressed in the Cartesian form 1/2+(1/2)i.
(b) Find the modulus of 4z − 2z*. (z* meaning z-bar/complex conjugate of z)
(c) Write 2z in polar form.
(d) Write 8z^3 in polar exponential form.

2. Relevant equations
(a) --
(b) =1+3i
(c) = 1+i, r = √2, θ=π/4
(d) = -2-2i

3. The attempt at a solution
(b) I ended up with 1+3i (is this right by the way?), with √1^2+3^2 = √10. So, does the modulus equal √10? However, I entered it into wolfram alpha and got 1-3i instead. So, which one is right?
(c) 2z=√2(cos(π/4)+isin(π/4)). -- Is that right? (Also with wolfram alpha, I got 2z=1-i (cartesian form) so, θ=-π/4 -- is this right instead?)
(d) I am confused with. How would I write out 8z^3? Would I first have to use it with (1/2)+(1/2)i and then write that given equation in polar exponential form? I tried that, with getting -2-2i and therefore, √8e^iπ/4. Is that correct?

Last edited: Apr 24, 2012
2. Apr 24, 2012

### HallsofIvy

For (a) it is very helpful to know that $i^2= -1$, $i^3= -i$, and $i^4= 1$. That means that powers of i are have "period 4". In particular, since 201= 50(4)+ 1, $i^{201}= i$; since 8= 2(4)+ 0, $i^8= 1$ so you are trying to find $z= -(1+ i)/(i(1+i)^2)= -1/(i(1+i)= i/(1+i)= (1+i)/2$ as claimed.
Then $\overline{z}= (1- i)/2$ so $4z- 2\overline{z}= 2+ 2i- 1+ i= 1+ 3i$ as you say. Are you sure you entered the problem into Wolfram alpha correctly?

Since $z= (\sqrt{2}/2)e^{i\pi/4}$, it follows that $z^3= (\sqrt{2}/2)^3e^{3i\pi/4}= (\sqrt{2}/4)e^{3i\pi/4}$. Multiplying by 8, $8z^3= 2\sqrt{2}e^{3i\pi/4}$. $\sqrt{8}= \sqrt{4(2)}= 2\sqrt{2}$ so that is correct but you appear to have forgotten that $(a^x)^3= a^{3x}$.

3. Apr 24, 2012

### Cottontails

I thought that my answers were right and I guess I must have somehow entered them incorrectly into wolfram alpha then. I was thinking about (2z)^3 = 8z^3 however, I didn't think you were able to rewrite the equation that way (some silly thinking, I know). Thank you for your help, that cleared out all the issues that I had with the question.