Express $(1-i)^{11}$ in cartesian form.
Apart from expanding the expression, I don't know how to do this. I've looked at the solution and still don't understand how/why it has been done.
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Hint: $1-i=\sqrt 2e^{-i\pi/4}{}{}{}{}$.
To find the above equality you can think about it geometrically and use the known techniques:
Or algebraically by proving that given any $a,b\in \mathbb R$, it holds that $a+ib=\sqrt{a^2+b^2}e^{i\arg(a+ib)}$, where $$\arg(a+ib)=\begin{cases} \arctan(b/a), &\text{if }a>0\\ \pi /2, &\text{if }a=0\land b>0\\ \arctan(b/a)+\pi, &\text{if }a<0\\ -\pi/2, &\text{if }a=0\land b<0\end{cases}$$
Git Gud
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A common trick is to factor the magnitude($\sqrt 2$) out of the expression. You should then recognize $e^{-iπ/4}$ – Gabriel Romon Jan 05 '14 at 12:11
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Hint: $(1-i)^2=-2i{}{}{}{}{}{}{}{}$.
Alraxite
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2I like this, it requires nothing more than basic complex algebraic manipulation. – Git Gud Jan 05 '14 at 12:20
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Expanding expressions like this one is not easy. Therefore, it is useful to rewrite it to its polar form. We know that the angle is $\varphi=-\frac 14 \pi$ and the length is $r=\sqrt 2$. We can find $r$ by Pythagoras' theorem: $$ r=\sqrt{1^2+(-1)^2}=\sqrt 2 $$ and we can find $\varphi$ by just looking at the point. (We know that $x=-y$, so the angle is of the form $k\frac 12\pi+\frac 14\pi$ for some integer $k$, and because it goes to the lower right, it has to be $-\frac 14\pi$.)
When taking the eleventh power, we get $$ \varphi'=11\varphi=-\frac{11}4\pi=-\frac 34\pi=\frac 54 \pi\\ r'=r^{11}\left(\sqrt2\right)^{11}=2^{\frac 12\cdot 11}=2^{5+\frac 12}=32\sqrt 2 $$ Now, we need to transform the result back to Carthesian form: $$ x=r'\cos\varphi'=-32\\ y=r'\sin\varphi'=-32 $$ Thus, we get the result $-32-32 i$ or $(-32,-32)$.
Ragnar
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@MrCroutini Plot the point $1-i$ or $(1,-1)$ in the plane; one unit east, one unit south of the origin. $\ \ r$ is the distance from $(1, -1)$ to $(0,0)$: $\sqrt((1)^2+(-1)^2)=\sqrt(2)$ by the distance formula. – bof Jan 05 '14 at 12:17
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$\theta$ is the angle the line from the origin to $1-i$ makes with the positive $x$-axis. Note that that line bisects the angle between the positive $x$-axis and the negative $y$-axis, so the angle is $-45^0$ or $-\pi/4$ radians. – bof Jan 05 '14 at 12:21
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Use Euler's Formula:
$(1-i)^{11} = \sqrt {2}^{11}*(\cos7\pi/4 + i\sin7\pi/4)^{11}$
= $\sqrt {2}^{11}*(\cos((7*11\pi)/4) + i\sin(7*11\pi)/4)$
= $\sqrt {2}^{11}*(-\frac{1}{\sqrt {2}}- i\frac{1}{\sqrt {2}})$
= $-32*(1+i)$
Thanks
Satish
Satish Ramanathan
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