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I do not know why I am having so many issues with complex numbers. I am basically trying to teach myself, but keep doubting myself and getting very frustrated. I've no way of checking my answers so I'm just lost. So I want to write the following in polar form $re^{i\theta}$ (although I thought that was called exponential form?) and $-\pi < \theta \leq \pi$.

My first question is expressing $(\cos(\frac{2\pi}{9}) + i\sin(\frac{2\pi}{9}))^3$. Using De Moivre, this is $$\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})$$ so it is $e^{i(\frac{2\pi}{3})}$ yes?

My next one is to express $\frac{2+2i}{-\sqrt{3} +i}$. So my attempt is to multiply by the complex conjugate and we get that this is $\frac{-\sqrt{3} +1}{2} + i\frac{-1 - \sqrt{3}}{2}$. Find $r$ which is the modulus which is $\sqrt2$ and $\theta$ is just something I cannot seem to find. I know if $z = a+ib$ then $\theta = \tan^{-1}(\frac{b}{a})$ but I'm stuck even with the $(\frac{b}{a})$ and am confusing myself of what quadrant to look in.

Lastly, I have no idea how to express the next one which is $\frac{4i}{3e^{(4+i)}}$.

Any help and direction at all is appreciated.

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3 Answers 3

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  • Yes, $\left(\cos\left(\frac{2\pi}9\right)+i\sin\left(\frac{2\pi}9\right)\right)^3=\left(\cos\left(\frac{2\pi}3\right)+i\sin\left(\frac{2\pi}3\right)\right)$.
  • Note that$$\frac{2+2i}{-\sqrt3+i}=\frac{1+i}{-\frac{\sqrt3}2+\frac12i}.$$Furthermore, $1+i=\sqrt2\left(\cos\left(\frac{\pi}4\right)+i\sin\left(\frac{\pi}4\right)\right)$ and $-\frac{\sqrt3}2+\frac12i=\cos\left(\frac{5\pi}6\right)+i\sin\left(\frac{5\pi}6\right)$. So, the quotient is equal to$$\sqrt2\left(\cos\left(\frac\pi4-\frac{5\pi}6\right)+i\sin\left(\frac\pi4-\frac{5\pi}6\right)\right)=\sqrt2\left(\cos\left(-\frac{7\pi}{12}\right)+i\sin\left(-\frac{7\pi}{12}\right)\right).$$
  • Note that$$\frac{4i}{3e^{4+i}}=\frac43\times\frac{e^{\frac{\pi i}2}}{e^{4+i}}=\frac43e^{-4+\frac{\pi i}2-i}.$$
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  • $\begingroup$ No, I am sorry. I am grateful for your help so far but I honestly have no idea where to go from there. I see/know how you rewrote the two complex numbers above but I do not know why or see how it helps in either case. $\endgroup$
    – moony
    Commented Feb 5, 2018 at 21:01
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    $\begingroup$ @moony I've edited my answer. Is it clear now? $\endgroup$ Commented Feb 5, 2018 at 21:06
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    $\begingroup$ @moony I'm using the fact that $\exp(x)/\exp(y)=\exp(x-y)$. $\endgroup$ Commented Feb 5, 2018 at 22:09
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    $\begingroup$ @moony It was an error. I've edited my answer. $\endgroup$ Commented Feb 5, 2018 at 22:19
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    $\begingroup$ @moony I'm glad I could help. $\endgroup$ Commented Feb 6, 2018 at 9:49
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You have $\frac {2+2i}{-\sqrt 3 + i}$ rather than rationalizing the denominator, and converting, convert numerator and denominator.

$2+2i = 2\sqrt2e^{\frac {\pi}{4}i}\\ -\sqrt 3 + i = 2e^{\frac {5\pi}{6}i}\\ \frac {2+2i}{-\sqrt 3 + i} = \sqrt 2\frac {e{\frac {\pi}{4}i}}{e{\frac {5\pi}{6}i}} = \sqrt2e^{(\frac {\pi}{4}i-\frac{5\pi}{6}i)}= \sqrt2e^{-\frac {7\pi}{12}i}$

Since you did get as far as:

$z = \frac {-\sqrt 3 + 1}{2} + \frac {-\sqrt 3 - 1}{2}i$

After doing it enough times in trig class I came to memorize $\cos \frac \pi{12} = \frac {\sqrt 6 + \sqrt 2}{4}, \sin \frac \pi{12} = \frac {\sqrt 6 - \sqrt 2}{4}$

Which might have helped.

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$\frac{2+2i}{-\sqrt{3} +i} =$

$\frac {(2+2i)(-\sqrt{3} -i)}{(-\sqrt{3} +i)(-\sqrt{3} -i)}=$

$\frac {(-2\sqrt{3} +2)- (2\sqrt 3+2)i}{3+1}=$

$\frac {-\sqrt3 + 1}2 -\frac{\sqrt 3 + 1}2i=$

$r(\cos \theta + i\sin \theta)$

So we need to solve for $r\cos \theta = \frac {-\sqrt3 + 1}2$ and $r\sin \theta = -\frac{\sqrt 3 + 1}2$.

That's all.

We solve for $r$ by squaring and adding both terms:

$r^2\cos^2 \theta + r^2 \sin^2 \theta = (\frac {-\sqrt3 + 1}2)^2 + ( -\frac{\sqrt 3 + 1}2)^2$

$r^2(\cos^2 \theta + \sin^2 \theta) = \frac {3 - 2\sqrt 3 + 1}4 + \frac {3+2\sqrt 3 + 1}4$.

$r^2 = 1 - \frac {\sqrt 3}2 + 1+\frac {\sqrt 3}2$

$r^2 = 2$

$r = \sqrt 2$.

And we solve for $\theta$ by dividing both terms and taking the arctangent:

$\tan \theta = \frac {\sin\theta}{\cos\theta} = \frac {r*\sin\theta}{r*\cos\theta} = \frac {-\frac{\sqrt 3 + 1}2}{\frac {-\sqrt3 + 1}2}$

$=\frac {\sqrt 3 + 1}{-\sqrt3 + 1} =\frac {(\sqrt 3 + 1)(-\sqrt 3- 1}{(-\sqrt3 + 1)}=\frac {-4 - 2\sqrt 3}{2} = -2-\sqrt 3$

So $\theta = \arctan -2-\sqrt 3= -\frac 5{12}\pi$

S $\frac{2+2i}{-\sqrt{3} +i} = \sqrt 2*e^{-\frac 5{12}\pi i}$.

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