Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question which should be super super easy!

If I was to solve $z^2 = 1+i$ how would I do this using de-moivres theorem?

I have the answer here in front of me so I don't want the answer, I just dont understand the method very well!

Any help would be appreciated! I haven't had much experience with complex numbers and have just started a complex analysis course.

Many thanks

share|improve this question
add comment

4 Answers

first write $1+i$ in polar form $\sqrt 2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$

Now, De-moivre's theorem says that $z=r(\cos\theta+i\sin\theta)\implies z^n=r^n(\cos n\theta+i\sin n\theta)$

which gives in your case $r^2(\cos2\theta+i\sin2\theta)=\sqrt 2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$

Now compare.

share|improve this answer
1  
You have some typos in your answer. –  Michael Hardy Mar 3 '13 at 21:25
add comment

$$ 1+i = \sqrt{2}\left(\cos\frac\pi4 + i\sin\frac\pi4\right) $$ It has two square roots: $$ \pm\sqrt{\sqrt{2}}\left(\cos\frac\pi8 + i\sin\frac\pi8 \right) = \pm 2^{1/4}\left(\cos\frac\pi8 + i\sin\frac\pi8 \right). $$

share|improve this answer
1  
I know the answer but how do you get that? –  camilla Mar 3 '13 at 21:41
add comment

$$\forall w=x+iy\in\Bbb C:$$ $$w=|w|e^{i\phi}=|w|(\cos\phi+i\sin\phi)\;,\;\;\phi=\begin{cases}\arctan\frac{y}{x}+2k\pi &,\;\;y\neq 0\\{}\\2k\pi\end{cases}\;\;,\;\;\;k\in\Bbb Z$$

In our case:

$$w=1+i\Longrightarrow |w|=\sqrt 2\,\,,\,\,\arctan\frac{1}{1}=\frac{\pi}{4}+2k\pi\Longrightarrow$$

$$z^2=1+i=\sqrt 2\,e^{\frac{\pi i}{4}\left(1+8k\right)}\;\;,\;\;k=0,1\;\Longrightarrow z=\sqrt[4]2\, e^{\frac{\pi i}{8}(1+8k)}\;,\;\;k=0,1$$

Why do we restrict ourselves only to the above values of $\,k\,$ ? Because any other integer value will give one of these two different ones (on the trigonometric circle, say) !

Thus, the solutions are

$$k=0:\;\;\;\;z_0:=\sqrt[4]2\,e^{\frac{\pi i}{8}}=\sqrt[4]2\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\\k=1:\;\;\;z_1:=\sqrt[4] 2\,e^{\frac{9\pi i }{8}}=\sqrt[4] 2\left(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8}\right)$$

By the way, it is a nice exercise to show that $\,z_0=-z_1\,$...

share|improve this answer
add comment

Let $1+i=r(\cos\theta+i\sin\theta)$ where $r\ge0,\theta $ are real

Equating the real & the imaginary parts, $r\cos\theta=1,r\sin\theta=1$

Squaring and adding we get $r^2(\cos^2\theta+\sin^2\theta)=1^2+1^2\implies r^2=2\implies r=\sqrt2$ as $r\not<0$

So, $\cos\theta=\frac1{\sqrt2}>0$ and $\sin\theta=\frac1{\sqrt2}>0\implies \theta=\frac\pi4$

So, $z^2=\sqrt2(\cos\frac\pi4+i\sin\frac\pi4)$

Using de Moivre's formula, $z=(\sqrt2)^\frac12\left(\cos\left(\frac{2r\pi+\frac\pi4}2\right)+i\sin\left(\frac{2rpi+\frac\pi4}2\right)\right)$ where $r=0,2-1=1$

For $r=0,z=2^\frac14\left(\cos\left(\frac{\frac\pi4}2\right)+i\sin\left(\frac{\frac\pi4}2\right)\right)=2^\frac14\left(\cos\frac\pi8+i\sin\frac\pi8\right)$

For $r=1,z=2^\frac14\left(\cos\left(\pi+\frac{\frac\pi4}2\right)+i\sin\left(\pi+\frac{\frac\pi4}2\right)\right)=-2^\frac14\left(\cos\frac\pi8+i\sin\frac\pi8\right)$ as $\cos(\pi+x)=-\cos x$ and $\sin(\pi+x)=-\sin x$

Of course, $r$ can assume any $2$ in-congruent values $\pmod 2$ to produce the two different roots.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.