# Thread: Roots and Complex Numbers

1. ## Roots and Complex Numbers

Hi folks

Got myself some dodgey questions to get through and Im bit clouded in what to do since the book did not help...

Find the real and imaginary parts of (1 + i)^35

Find all complex solutions to the equation (z + 1)^4 = 16z^4

And

Find the complex logarithm of [sqrt]3 + i, find a complex number z=a+bi (where a and b are real), such that e^z = [sqrt]3 +i

Any help is appreciated

2. Originally Posted by NeloAngelo
Find the real and imaginary parts of (1 + i)^35
$\displaystyle 1 + i = \left| {1 + i} \right|e^{i\arg (1 + i)} = \sqrt 2 e^{i\frac{\pi } {4}}$

therefore:

$\displaystyle (1 + i)^{35} = 2^{\frac{{35}} {2}} e^{i\frac{{35\pi }} {4}}$

$\displaystyle \Re e\left\{ {(1 + i)^{35} } \right\} = \Re e\left\{ {2^{\frac{{35}} {2}} e^{i\frac{{35\pi }} {4}} } \right\} = 2^{\frac{{35}} {2}} \cos \left( {\frac{{35\pi }} {4}} \right)$

$\displaystyle \Im m\left\{ {(1 + i)^{35} } \right\} = \Im m\left\{ {2^{\frac{{35}} {2}} e^{i\frac{{35\pi }} {4}} } \right\} = 2^{\frac{{35}} {2}} \sin \left( {\frac{{35\pi }} {4}} \right)$

3. Originally Posted by NeloAngelo
Hi folks
Find all complex solutions to the equation (z + 1)^4 = 16z^4
$\displaystyle \begin{gathered} \left( {z + 1} \right)^4 = 16z^4 \hfill \\ \Leftrightarrow \left( {z + 1} \right)^4 = \left( {2z} \right)^4 \hfill \\ \Leftrightarrow \left( {z + 1} \right)^4 - \left( {2z} \right)^4 = 0 \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} \Leftrightarrow \left[ {\left( {z + 1} \right)^2 - \left( {2z} \right)^2 } \right]\left[ {\left( {z + 1} \right)^2 + \left( {2z} \right)^2 } \right] = 0 \hfill \\ \Leftrightarrow \left( {3z + 1} \right)\left( {1 - z} \right)\left( {5z^2 + 2z + 1} \right) = 0 \hfill \\ \end{gathered}$

can you continue?

4. Hi Thx for the feedback
I presume when the question states Complex Solutions they mean Complex Roots therefore after using the binomial expansion I use the formula involving arg, sin and cos. Is that correct?

5. Originally Posted by NeloAngelo
Hi Thx for the feedback
I presume when the question states Complex Solutions they mean Complex Roots therefore after using the binomial expansion I use the formula involving arg, sin and cos. Is that correct?
Where do you figure the binomial expansion gets used?? Clearly not in the two Peritus has answered. And not in the one I'm about to help with .....

Originally Posted by NeloAngelo
[snip]
Find the complex logarithm of [sqrt]3 + i, find a complex number z=a+bi (where a and b are real), such that e^z = [sqrt]3 +i

Any help is appreciated
You should know that $\displaystyle \ln w = \ln |w| + i \theta$ where $\displaystyle \theta = \arg w$ ......

Now note that $\displaystyle \ln (\sqrt{3} + i) = z \Rightarrow \sqrt{3} + i = e^z$ .....

6. Sorry my bad. So is it just the quadratic equations I got to use along with Natural Logs and de Moivre Equation?