Calculating real and imaginary part of a complex number

Question

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.

Answer 1

Try to understand and prove each step:

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

Answer 2

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 $\Im(z)=\frac{z-z^*}{2i}$.

Answer 3

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 $$ e^{b-i\pi}=e^be^{-i\pi}=e^b(\cos\pi-i\sin\pi) $$

Answer 4

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.