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Consider the complex numbers $a = \frac{(1+i)^5}{(1-i)^3}$ and $b = e^{3-\pi i}$.

How do I calculate the real and imaginary part of these numbers? What is the general approach to calculate these parts?

I thought about reforming them to the form $x + i\cdot y$ which might be possible for a, but what about b?

I just started occupying with complex numbers and don't yet understand the whole context.

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4 Answers 4

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Try to understand and prove each step:

$$\begin{align*}\bullet&\;\;\frac{1+i}{1-i}=i\implies \frac{(1+i)^5}{(1-i)^3}=\left(\frac{1+i}{1-i}\right)^3(1+i)^2=i^3\cdot2i=(-i)(2i)=2\\{}\\\bullet&\;\;e^{b-\pi i}=e^be^{-\pi i}=e^b\left(\cos\pi-i\sin\pi\right)=-e^b\end{align*}$$

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  • $\begingroup$ Just wondering since you said prove each step does $e^{b-\pi i}=e^be^{-\pi i}$ need to be proved(is there a catch apart from the usual law of exponents) $\endgroup$
    – Guy
    Commented Apr 21, 2014 at 16:11
  • $\begingroup$ Not really in this case as the exponent is definitely determined (of course, $\;b\in\Bbb R\;$ here), so we don't have to hesitate between different arguments... $\endgroup$
    – DonAntonio
    Commented Apr 21, 2014 at 16:13
  • $\begingroup$ Okay thanks. :) $\endgroup$
    – Guy
    Commented Apr 21, 2014 at 16:13
  • $\begingroup$ thank you for your answer. The first one is nearly clear to me, except for $\frac{1+i}{1-i} = i$, could you please help me with this one? For the 2nd one: $e^{b-\pi i}$ could be (afaik) turned into $\frac{e^b}{e^{\pi i}}$, but how do you get to $e^b e^{-\pi i}$? $\endgroup$
    – muffel
    Commented Apr 21, 2014 at 16:19
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    $\begingroup$ @muffel, multiply by the denominator's conjugate, as usual:$$\frac{1+i}{1-i}\frac{1+i}{1+i}=\frac{2i}2=i$$ About the second one: isn't $\;e^{-c}=\frac1{e^c}\;\;?$ . Just the same thing in two different dresses... $\endgroup$
    – DonAntonio
    Commented Apr 21, 2014 at 16:22
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A useful way is the trigonometric one:

$a=\frac{(1+i)^5}{(1-i)^3}$.

Observe that \begin{align*}1+i=&\frac{2}{\sqrt2}\left(\frac{\sqrt2}{2}1+\frac{\sqrt2}{2}i\right)\\ =&\frac{2}{\sqrt2}(\cos(\pi/4)+i\sin(\pi/4))\\ =&\frac{2}{\sqrt2}e^{\frac{i\pi}{4}}.\end{align*}

Then $$ 1-i=\overline{1+i}=\overline{\frac{2}{\sqrt2}e^{\frac{i\pi}{4}}}=\frac{2}{\sqrt2}e^{\frac{-i\pi}{4}}. $$

Hence \begin{align*} a=&\frac{(1+i)^5}{(1-i)^3}\\ =&(1+i)^5(1-i)^{-3}\\ =&\left(\frac{2}{\sqrt2}e^{\frac{i\pi}{4}} \right)^5\left(\frac{2}{\sqrt2}e^{\frac{-i\pi}{4}}\right)^{-3}\\ =&\frac{2^2}{2}e^{2i\pi}=2\;. \end{align*}

Finally $b=e^{3-\pi i}=e^3e^{-\pi i}=e^3(\cos(-\pi)+i\sin(-\pi))=-e^3$ hence $\Re b=-e^3$ and $\Im b=0$.

However the approach depends on the case you face.

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  • $\begingroup$ thank you for your answer. Could you please explain to me how you got from $1+i$ to $\frac{2}{\sqrt{2}}\left(\dots\right)$? $\endgroup$
    – muffel
    Commented Apr 21, 2014 at 16:23
  • $\begingroup$ Basically experience: you do a lot of complex numbers exercises in order to "form your eye". $z=1+i$ is a number in which $\Re z=\Im z$ and then I asked myself: "what is the value of $\alpha$ that makes $\cos\alpha=\sin\alpha$" (wlog $\alpha\in[0,2\pi[$)? It's $\alpha=\pi/4$ (even $\alpha=\frac{5\pi}{4}$), but $z$ and $\cos\alpha+i\sin\alpha$ differs by a multiplicative constant, that is $|z|=\sqrt{1^2+1^2}=\sqrt2=\frac{2}{\sqrt2}$. Then I adjusted all by using this constant... and this is all! $\endgroup$
    – Joe
    Commented Apr 21, 2014 at 16:53
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Consider any complex number $z=x+iy$, where $x$ and $y$ are the real and imaginary parts, respectively. The complex conjugate of $z$ will be $z^*=x-iy$. Note that if we add $z$ and $z^*$ together we get $z+z^*=2x$, so the real part of $z$ may be written as $\Re(z)=\frac{z+z^*}{2}$. This allows to circumvent decomposing a complex number into the form $x+iy$ in order to find the real part. All you need to do is compute the complex conjugate. You can similarly find that the imaginary part of $z$ can be written as $\Im(z)=\frac{z-z^*}{2i}$.

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HINT : $$ \begin{align} a=\frac{(1+i)^5}{(1-i)^3}&=\frac{(1+i)^3(1+i)^2}{(1-i)^3}\\ &=\left(\frac{1+i}{1-i}\right)^3(1+i)^2\\ &=\left(\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}\right)^3(1+2i-1)\\ &=i^3\cdot2i \end{align} $$ and $$ e^{b-i\pi}=e^be^{-i\pi}=e^b(\cos\pi-i\sin\pi) $$

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