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Write the complex number in polar form:

$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$

So my try goes as follows:

\begin{align} \frac{(1−i)^3(\sqrt 3+i)}{4i} &= \frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& = \frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& = \frac{(-2-2i)(\sqrt 3+i)\times-4i}{16} \\&= \frac{(-1-i)(\sqrt 3+i)\times-4i}{8}\\& = \frac{(4i - 4)(\sqrt 3+i)}{8} \\&= \frac{(i-1)(\sqrt 3+i)}{2} \\&= \frac{(\sqrt 3 \times i-1-\sqrt 3-i)}{2}\\& = \frac{-(\sqrt 3+1)}{2} + \frac{(\sqrt 3-1)i} {2}\end{align}

Polar form:

$$r =\sqrt {\left(\frac {-1+\sqrt 3)}{2}\right)^2 + \left(\frac{\sqrt 3-1}{2}\right)^2} = \sqrt{2} $$

$$\tan(v) = \frac{(\sqrt 3-1)} { -(1+\sqrt 3)} \iff v = \arctan(\sqrt 3-2) $$ put it equal to $K$

$$z = \sqrt{2}(\cos K, i \sin K), \\ K = \arctan(\sqrt 3-2)$$

To me, this doesn’t seem like a clean answer, since this question could potentially be on an upcoming exam. Is there something I am missing, like a better approach?

Rócherz
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The anime kid
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    You could split it into three complex numbers $ 1-i, 4, \sqrt{3}+i$ And find the polar form (absolute value and arg) of each and then use that – bananapeel22 Oct 18 '22 at 14:24
  • No need to compute the polar form of the entire expression. Since it's highly multiplicative you can compute the polar form of each of the 3 elementary parts and use the product/quotient rules for modulus and argument. – Lelouch Oct 18 '22 at 14:56

2 Answers2

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The beat way to change to polar form is to change individual terms to polar form first So $$(1-i)^3=\left(\sqrt{2}e^{-\frac{\pi i}{4}}\right)^3=2\sqrt{2}e^{-\frac{3\pi i}{4}}$$ $$\sqrt{3}+i=2e^{\frac{\pi i}{6}}$$ and $$4i=4e^{\frac{\pi i}{2}}$$

Which means that $$z=\sqrt{2}e^{-\frac{13\pi i}{12}}$$

Sam
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    Note: when using fractions in exponents, you can use \tfrac instead of \frac to get something less tiny e.g. $e^{-\tfrac{13\pi}{12}}$, there exists also \dfrac but this is then too large. – zwim Oct 18 '22 at 15:18
  • Thank you! Will be sure to use it in my next posts. That’s very helpful – Sam Oct 18 '22 at 15:28
  • Note: You can also write the $e^{-\frac{3}{4} \cdot \pi \cdot \mathrm{i}}$-stuff as $e^{-\frac{3}{4} \cdot \pi \cdot \mathrm{i}} = \exp\left(-\frac{3}{4} \cdot \pi \cdot \mathrm{i}\right) = \operatorname{cis}\left(-\frac{3}{4} \cdot \pi\right)$. Then you don't have the exponent that matters so small in the corner... Then it's easier to read. ;) – Kevin Dietrich Oct 18 '22 at 15:43
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Problems

The absolute value $r$ is correct, but the angle is not! You used the wrong formula for the angle...

If you got a complex number $z = a + b \cdot \mathrm{i} = r \cdot \operatorname{cis}(\theta)$ then your angle is $\theta = \operatorname{arctan2}(b, ~a)$. But you used another formula.

If you use $\theta = \operatorname{arctan2}(b, ~a)$ you will get: $\theta = \operatorname{arctan2}\left(\Im\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right), ~\Re\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right)\right) = \arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi = \frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi$.

better way to approach this

Is there something I am missing, like a better way to approach this?

Yes there is.

But you could also write the individual terms that are multiplied or divided with each other in polar form and then summarize them using the power laws.

If you don't whant to this 'cause you don't like the polar form. You just can say/use (algebraic): $$ \begin{align*} z &= a + b \cdot \mathrm{i}\\ z &= r \cdot \operatorname{cis}(\theta)\\ \\ r &= |z| = \sqrt{a^{2} + b^{2}}\\ \theta &= \arg(z) = \operatorname{arctan2}\left( b, ~a \right)\\ \\&\text{with}\\\\ \operatorname{arctan2}\left( b, ~a \right) &= \begin{cases} \arctan\left({\frac {b}{a}}\right) \qquad\quad~~~ \text{for } a > 0\\ \arctan\left({\frac {b}{a}}\right) + \pi \qquad \text{for } a < 0 \quad y > 0\\ \pi \qquad\qquad\qquad\quad~~~ \text{for } a < 0 \quad y = 0\\ \arctan\left({\frac {b}{a}}\right) - \pi \qquad \text{for } a < 0 \quad y < 0\\ \frac{\pi}{2} \qquad\qquad\qquad\quad~~ \text{for } a = 0 \quad y > 0\\ -\frac{\pi}{2} \qquad\qquad\qquad~~~ \text{for } a = 0 \quad y < 0\\ \end{cases} + 2 \cdot k \cdot \pi \end{align*} $$

If you use that you'll get: $$ \begin{align*} z &= \frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\\ z &= -\frac{1 + \sqrt{3}}{2} + \left( \frac{-1 + \sqrt{3}}{2} \right) \cdot \mathrm{i}\\ z &= \sqrt{\left( -\frac{1 + \sqrt{3}}{2} \right)^{2} + \left( \frac{-1 + \sqrt{3}}{2} \right)^{2}} \cdot \operatorname{cis}\left(\operatorname{arctan2}\left( -\frac{1 + \sqrt{3}}{2}, ~\frac{-1 + \sqrt{3}}{2} \right)\right)\\ z &= \sqrt{2} \cdot \operatorname{cis}\left(\arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi\right)\\ z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi\right)\\ z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi\right)\\ z &= \sqrt{2} \cdot \left(\cos\left(\frac{11}{12} \cdot \pi\right) + \sin\left(\frac{11}{12} \cdot \pi\right) \cdot \mathrm{i} \right)\\ \end{align*} $$

Kevin Dietrich
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