2
$\begingroup$

I'm going through past exam papers and have come accross the following question:

find all solutions to $z^5=2-2i $

for this question I was going to use Euler's formula:

$ e^{i(2k+1)\pi}=-1 $

$ -2e^{i(2k+1)\pi}=2 $

$ -2e^{i(2k+1)\pi}-2i=2-2i $

$ z^5 =-2e^{i(2k+1)\pi}-2i$

$ z = -2^{\frac 1 5}e^{\frac {i(2k+1)\pi}{5}}-2i^{\frac 1 5}$

I'll then sub in k=0,1,2,3,4

am I on the right track here? thanks!

  • 0
    Try writing $2-2i$ in the $e^{x}$ form first – 2017-01-05
  • 2
    $(a+b)^{1/5}\neq a^{1/5}+b^{1/5}$ – 2017-01-05

4 Answers 4

0

Using the Euler's Formula is a good approach but I would use:

$$z=re^{ix} \rightarrow z^5=r^5e^{5xi}$$

$$2-2i=2\sqrt{2}e^{(7\pi/4+2k\pi)i}$$

Now compare both:

$$r^5e^{5xi}=2\sqrt{2}e^{(7\pi/4+2k\pi)i}$$

Once both complex numbers are equal then they must have the same magnitude ($r^5=2\sqrt{2}$) and also $e^{5xi}=e^{(7\pi/4+2k\pi)i}$, so:

\begin{cases} r^5=2\sqrt{2}\\ e^{5xi}=e^{(7\pi/4+2k\pi)i} \end{cases}

P.S: Remember that in general if we have $u=r_1e^{ix_1}$\ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.

Can you finish?

  • 0
    just confused at where you got $ e^{5xi}=e^{(7\pi/4+2k\pi)i} $ Thanks for your reply! – 2017-01-05
  • 0
    I'm just comparing the Euler formula. If we have $z=r_ze^{xi}$ and $w=r_we^{yi}$ then if $z=w$ that give us both numbers has the same modulus ($r_z=r_w$) and also $e^{xi}=e^{yi}$. – 2017-01-05
  • 0
    @user6122081: are you ok now? – 2017-01-05
0

We can write $2-2i$ as:

$z^5=2-2i=2\sqrt2e^{i{-\pi\over4}}=2\sqrt2e^{i{-\pi\over4}+2k\pi}$

now we take the root:

$z=(2\sqrt2)^{1\over5}e^{i{-\pi\over20}+{2k\pi\over5}}$

with $k\in \mathbb Z$, $k=0,1,2,3,4$.

Your approach fails because $z=(2-2i)^{1\over5}\ne(2)^{1\over5}-(2i)^{1\over5}$

  • 0
    $arg(2-2i)$ is $\displaystyle7\frac{\pi}{4}$ – 2017-01-05
0

Hint:

$$e^{7\pi/4}=\frac1{\sqrt 2}-i\frac1{\sqrt 2}$$

Therefore

$$2-2i=2\sqrt 2e^{7\pi/4}$$

0

You fell for the freshman's dream: $(a+b)^n=a^n+b^n$.

So, instead, write $2-2i=2\sqrt2e^{\frac{7\pi}4i}$ (using Euler's formula).

Now $z^5=2\sqrt2e^{\frac{7\pi}4i}$.

So $z=\sqrt[10]{8}e^{\frac{7\pi}{20}i}$ is one solution.

There are $4$ more.

Take a primitive $5$th root of unity, $\rho=e^{\frac{2\pi i}5}$. Then the other $4$ are: $\rho z,\rho^2z,\rho^3z$ and $\rho^4z$.

That is, $\sqrt[10]8e^{(\frac{7\pi+8\pi k}{20})i}$, for $k=0,1,2,3$ and $4$.