Results 1 to 2 of 2

Thread: complex simplifying

  1. #1
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432

    complex simplifying

    Simplify $\displaystyle \sqrt{15+8i}$.

    I got $\displaystyle 4+i$ or $\displaystyle -4 - i$, but only one of these solutions is correct.

    Obviously $\displaystyle \sqrt{15+8i}>0$, but with my answers, wouldn't the pesky $\displaystyle i$ term disable you from checking whether the expression is greater or less than 0? After all, $\displaystyle i$ is not on the number line... I dunno, it just feels risky dealing with that $\displaystyle i$ term.

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by DivideBy0 View Post
    Simplify $\displaystyle \sqrt{15+8i}$.

    I got $\displaystyle 4+i$ or $\displaystyle -4 - i$, but only one of these solutions is correct.

    Obviously $\displaystyle \sqrt{15+8i}>0$, but with my answers, wouldn't the pesky $\displaystyle i$ term disable you from checking whether the expression is greater or less than 0? After all, $\displaystyle i$ is not on the number line... I dunno, it just feels risky dealing with that $\displaystyle i$ term.

    thanks
    Given any (non-zero) number $\displaystyle z$ its square root $\displaystyle \sqrt{z}$ is $\displaystyle \sqrt{|z|} \cdot e^{\theta i/2}$ where $\displaystyle \theta$ is the angle it makes with the x-axis measure $\displaystyle (-\pi,\pi]$.*

    So if $\displaystyle z=15+8i$ then $\displaystyle |z| = \sqrt{15^2+8^2} = \sqrt{298}$.

    The angle that $\displaystyle 15+8i$ makes with the x-axis is $\displaystyle \theta = \tan^{-1} (15/8)$.

    So,
    $\displaystyle e^{i\cdot \left( \frac{\tan^{-1}(15/8)}{2} \right) } = \cos \left( \frac{\tan^{-1}(15/8)}{2} \right) + i \sin \left( \frac{\tan^{-1} (15/8)}{2} \right) $

    Use half-angle formulas,
    $\displaystyle \frac{1}{\sqrt{2}} \sqrt{ 1 + \cos \tan^{-1}(15/8) } + i\frac{1}{\sqrt{2}} \sqrt{ 1 - \cos \tan^{-1}(15/8)}$

    Can you simplify that?

    *)And this is computed with Euler formula: $\displaystyle e^{xi} = \cos x + i\sin x$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplifying complex numbers?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Apr 24th 2011, 05:14 PM
  2. Simplifying a complex rational
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Nov 14th 2010, 07:54 PM
  3. Simplifying complex fraction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Oct 30th 2010, 09:11 PM
  4. Simplifying Complex Numbers (i)
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Mar 18th 2008, 09:17 PM
  5. simplifying a complex fraction
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Oct 18th 2007, 07:28 AM

Search Tags


/mathhelpforum @mathhelpforum