Algebra - Complex Numbers

Calculate the length of the vector given by $\frac{\left( 1+2i \right)\left( 1+\sqrt{3}\cdot i \right)}{\left( 1+i \right)^{3}} $ I start by doing $\left( 1+i \right)^{3}=2\left( -1+i \right)$

And then I multiply both the numerator and the denominator with $\left( -1-i \right)$

See my calculation:

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I can't spot where I'm going wrong in my calculation.


1 Answer 1


Alternatively you can rewrite your number in its polar form:

  • $1+2i=\sqrt 5e^{i\theta _1}$, for some real $\theta_1$;
  • $(1+\sqrt 3i)=2e^{i\theta _2}$, for some real $\theta_2$;
  • $(1+i)^3=(\sqrt 2e^{i\theta _3})^3$, for some real $\theta _3$.

Yielding $\displaystyle \left\vert\dfrac{\left( 1+2i \right)\left( 1+\sqrt{3} i \right)}{\left( 1+i \right)^{3}}\right \vert=\left \vert\dfrac{\sqrt 5e^{i\theta _1}\cdot2^{i\theta _2}}{2\sqrt 2e^{3\theta _3 i}}\right \vert=\sqrt{\dfrac{5}{2}}.$

In your work you get $\dfrac{1}{2(-1+i)}=\dfrac{-1-i}{2}$ which is wrong.
Correct is $\dfrac{1}{2(-1+i)}=\dfrac{-1-i}{4}$.

This still doesn't excuse for the absolute value of what you get being negative, but that becomes unimportant.

In any case you squared $2\sqrt 3-\sqrt 3$ wrong: $$4\cdot 3-4\cdot 3+3=(2\sqrt 3-\sqrt 3)~2\neq 4\color{red}{\sqrt 3}-4\cdot 3+3.$$

  • 1
    $\begingroup$ This not only tells how it can be done; it also actually answers the question where the error is. (+1) $\endgroup$
    – Mårten W
    Sep 15, 2013 at 13:00
  • $\begingroup$ excellent answer, thanks. $\endgroup$
    – jacob
    Sep 22, 2013 at 12:56
  • $\begingroup$ @jacob You're welcome. $\endgroup$
    – Git Gud
    Sep 22, 2013 at 14:07

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