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

  1. Apr 24, 2012 #1
    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
    (a) I have already done.
    (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. jcsd
  3. Apr 24, 2012 #2


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

    Since [itex]z= (\sqrt{2}/2)e^{i\pi/4}[/itex], it follows that [itex]z^3= (\sqrt{2}/2)^3e^{3i\pi/4}= (\sqrt{2}/4)e^{3i\pi/4}[/itex]. Multiplying by 8, [itex]8z^3= 2\sqrt{2}e^{3i\pi/4}[/itex]. [itex]\sqrt{8}= \sqrt{4(2)}= 2\sqrt{2}[/itex] so that is correct but you appear to have forgotten that [itex](a^x)^3= a^{3x}[/itex].
  4. Apr 24, 2012 #3
    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.
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