# solving a basic complex equation but using de Moivres theorem

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

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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.

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You have some typos in your answer. – Michael Hardy Mar 3 '13 at 21:25

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

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I know the answer but how do you get that? – camilla Mar 3 '13 at 21:41

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

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 $\,z_0=-z_1\,$...

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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.

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