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Find the cartestian form of $6 - 7i$ rotated anticlockwise through $\frac{3\pi}{4}$ about the origin

I realize that I am going to be doing something like:

$\sqrt{85}e^{i\alpha}.e^{i\frac{3\pi}{4}}$ where $\alpha = \arctan{(\frac{7}{6})}$

and then converting to Cartesian form. I guess I am having trouble dealing with the $\alpha$ term since the answer I am trying to arrive at is $\frac{1 + 13i}{\sqrt{2}}$ , i.e, an exact representation not involving any $\arctan$ terms.

How do I arrive at the given answer?

2 Answers 2

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You do not need to convert 6-7i to polar form at all. Geometry suffices to show us $e^{i\pi/4}=\frac{-1+i}{\sqrt2}$; simply multiply your original complex number by this in standard $a+bi$ form.

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    An elegant solution!2012-04-01
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You should have

$ r(cos(\theta) + isin(\theta))$

Your problem seems to be that $\theta = arctan(7/6) + \frac{3\pi}{4}$

You might try using the compound angle formulae:

$sin( arctan(\frac{7}{6}) + \frac{3\pi}{4}))$

$= sin(arctan(\frac{7}{6}))cos(\frac{3\pi}{4})) + sin(\frac{3\pi}{4}))cos(arctan(\frac{7}{6}))$

We can use a right-angled triangle idea (we can find the sign by noting that we're in the 1st quadrant) to find $sin(arctan(\frac{7}{6}))$. If we create a right-triangle with opposite 7 and adjacent 6, it's hypotenuse will be $\sqrt(85)$, so:

  • $sin(arctan(\frac{7}{6}))$ = $\frac{7}{\sqrt{85}}$

  • $cos(arctan(\frac{7}{6}))$ = $\frac{6}{\sqrt{85}}$

This should get you to the answer quickly.

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    I'm not sure if the ambiguity of negative signs is dealt with well enough? But @robjohn's answer is a better solution anyway.2012-04-01