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

Question

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?

Answer 1

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.

Answer 2

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.