Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 14

MATH152 April 2013

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Question A 14
Let $z={\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}i$. Express the complex number $z^{2010}$ in the form $a+bi$ where a and b are real numbers.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use Euler's formula, which states that $e^{ix}=\cos(x)+i\sin(x)$ and find the value of x that gives the given complex number. then take powers.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Euler's formula states that for a complex number given in argument form, we have $\displaystyle \cos(\theta )+i\sin(\theta )=e^{i\theta }$ In the case of the complex number $z={\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}i$ we can either see that this corresponds to the point at distance 1 of the origin on the diagonal in the first quadrant of the complex plan and hence has an argument of 45° or of π/4; or you might directly remember the values of the sine and cosine of π/4. Either way, we obtain that $z={\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}i=\cos(\pi /4)+i\sin(\pi /4)=e^{i\pi /4}$ Taking the 2010 of $z$ is much easier in this shape. We have $\displaystyle z^{2010}=(e^{i\pi /4})^{2010}=e^{i1005\pi /2}$ In this notation we see that the argument of $z^{2010}$ is $1005\pi /2$ which remains the same if we remove or add $4\pi /2$, so since $2005=4\cdot 501+1$ we have that the angle $1005\pi /2=\pi /2$ and so Using Euler's formula again yields $\displaystyle z^{2010}=e^{(1005)i\pi /2}=e^{i\pi /2}$ Using Euler's formula the other way around we obtain that $\displaystyle z^{2010}=e^{i\pi /2}=\cos(\pi /2)+i\sin(\pi /2)=i$ Since $\cos(\pi /2)=0$ and $\sin(\pi /2)=1$.

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