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Currently studying for my calculus exam when I stumbled upon this example:

Find the real and imaginary parts of the following

$$z=(1+i)^{100}$$

Following the answer to this problem it's first stated that $1+i$ can be written in polar form as $\sqrt{2}e^{i\frac{\pi }{4}}$ which I totally get. Proceeding by rewriting it as:

$$z=\sqrt{2}^{100} * (e^{i\frac{\pi}{4}})^{100}$$

Then we get to the parts where I'm totally lost at the moment.

$$2^{50}*e^{i2\pi} = 2^{50}*e^{i\pi} = -2^{50}$$

Where the real part would be equal to $-2^{50}$ and the imaginary part to be equal to $0$ which I can see in the answer given.

However, the last line of simplification is what confuses me. Could someone explain what is done?

Thanks!

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    $\begingroup$ Not a duplicate since the OP is asking for an explanation in a step of his proof and the linked question is asking for alternative methods to this problem. $\endgroup$ – vrugtehagel Mar 13 '17 at 17:14
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    $\begingroup$ note that $$-2^{50}$$ is the right answer $\endgroup$ – Dr. Sonnhard Graubner Mar 13 '17 at 17:15
  • $\begingroup$ @vrugtehagel You're absolutely right, thanks for pointing that out. $\endgroup$ – user409521 Mar 13 '17 at 17:20
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It looks like there was a typo:

$$\sqrt{2}^{100} \cdot \left(e^{\tfrac\pi4i}\right)^{100} = 2^{50} \cdot e^{25 \pi i} = 2^{50} \cdot e^{\pi i} = -2^{50}$$

The first equality is just multiplying powers, and they had $2\pi$ instead of $25\pi$. The second equality comes from the fact that $e^{2\pi i} = 1$. The third equality uses $e^{\pi i} = -1$

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  • $\begingroup$ Yeah, it's probably a typo as you say. $e^{i\pi } = -1$ is because of the equality $e^{i\pi }=\cos \left(\pi \right)+i\sin \left(\pi \right)$, correct? $\endgroup$ – SmhConfused Mar 13 '17 at 17:36
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    $\begingroup$ I suspect they left out the $5$ accidentally, and intended it to be $25\pi$ instead of $2\pi$. Otherwise, their stated equality is actually not true. Yes, $e^{i\pi} = \cos(\pi) + i\sin(\pi)$ will verify that $e^{i\pi} = -1$. $\sin(\pi) = 0$, and $\cos(\pi) = -1$. The same formula will verify $e^{2\pi i} = 1$. $\endgroup$ – Christian Mar 13 '17 at 17:54
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What about if we start with $$(1+i)^2=1+2i-1=2i.$$ Then $$(1+i)^{100}=[(1+i)^2]^{50}=(2i)^{50}=2^{50}(i^2)^{25}=2^{50}(-1)^{25}=-2^{50}.$$ So the real part is $-2^{50}$ and the imaginary part is $0$.

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I believe the 'part where you are lost' should read:

$2^{50}∗e^{25*iπ}$ = $2^{50}∗e^{6*4iπ}∗e^{iπ}$ = $2^{50}∗e^{iπ}$ = $2^{-50}$

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    $\begingroup$ There's a mistake there: $\;2^{50}\cdot e^{i\pi}=2^{50}\cdot(-1)=-2^{50}\neq 2^{-50}\;...$ $\endgroup$ – DonAntonio Mar 13 '17 at 17:29
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With a similar starting point to a given answer, but a different pathway:

$$(1+i)^2 = 2i$$

and squaring this

$$(1+i)^4 = -4 = - 2^2$$

and then

$$((1+i)^4)^{25} = (1+i)^{100} = (-)^{25} (2^{2\times25}) = -2^{50}$$ as before.

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The squere root can be written as

\begin{equation} \sqrt{x} = x^{\frac{1}{2}} \end{equation} So \begin{equation} \sqrt{2}^{100} = (2^{\frac{1}{2}})^{100}=2^{50} \end{equation} While \begin{equation} (e^{j \frac{\pi}{4}})^{100} = e^{j 25 \pi}= e^{j \pi} e^{j 24 \pi}=-1 e^{j 24 \pi} = -1 \end{equation}

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