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So we have:

z = 7 + 4i

w = 8-i

I did this: $$7+4i/8-i$$ $$conjagate =\space 8+i$$ $$ (7+4i/8-i) \times (8+i/8+i)$$ $$(56+7i+32i+4i^2)/((8-i)\times(8+i))$$ $$52+39i/(65)$$

This is what I got using polar form:

Z in polar form

$$Polar form = x+iy = r(cos(0)+isin(0))$$ $$ r = \sqrt{x^2+y^2}$$


$$r (for\space z) = \sqrt{7^2+4^2} = 2072/257$$

$$ 0 = atan(y/x) = atan(4/7) =14/27$$

therefore z in polar form is


W in polar form

$$Polar form = x+iy = r(cos(0)+isin(0))$$ $$ r = \sqrt{x^2+y^2}$$


$$r (for\space w) = \sqrt{8^2+-1^2} = 2024/255$$

$$ 0 = atan(y/x) = atan(-1/8) =-241/1938$$

therefore w in polar form is

$$(2024/255)*(cos(-241/1938)+i*sin(-241/1938)) = w$$

$$z/w = (\sqrt(7^2+4^2)*(cos(atan(4/7))+i*sin(atan(4/7)))) / (\sqrt(8^2+(-1)^2)*(cos(atan((-1)/8))+i*sin(atan((-1)/8))))$$

$$z/w = 4/5+3/5i$$

All good, it works out!

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They are not the same because you made a different mistake the first time from the mistake you made the second time. – Gerry Myerson Aug 8 '13 at 13:15
What did I mess up in the first part, I just checked my equations and they seem fine? – Ghozt Aug 8 '13 at 13:16
Please show explicitly the steps you took while computing because you got wrong answers in both parts. – Ruslan Aug 8 '13 at 13:18
I have no idea what you've done, really; but your numbers in polar are not the same as in rectangular. Though they are very close. – FireGarden Aug 8 '13 at 13:19
In the first part, there is no need for parentheses around the $65$. There is need for parentheses around $52+39i,$ for if not, then you're talking about $$52+\frac{39i}{65}.$$ As for the second part, you're doing fine, except you keep approximating. For $z$, $$r=\sqrt{55}\neq\frac{2072}{257}$$ and $$\theta=\arctan\frac47\neq\frac{14}{27}.$$ – Cameron Buie Aug 8 '13 at 13:41

So first off:

$$\frac {7+4i}{8-i} \times \frac{8+i}{8+i}=\frac {52+39i}{65}=.8+.6i$$ So In polar form this is $e^{arctan(\frac34)}$

Suppose we have a complex variable defined as $a+ib$. Then in polar coordinates, this is equivalent to $re^{\theta}$, where $r=\sqrt {a^2+b^2} ; \theta = tan^{-1}(\frac ba).$

So we have

z = 7 + 4i

w = 8-i

In polar form, z = $\sqrt{4^2+7^2}e^{arctan{\frac 4 7}}=\sqrt{65}e^{arctan(\frac 47)}$

And w = $\sqrt {8^2+1^2}e^{arctan(-\frac 18)}=\sqrt{65}e^{-arctan (\frac18)}$

So $\frac zw = e^{arctan(\frac 47)+arctan(\frac 18)}$

If you do a little arithmetic, you see that the above exponent is equal to $e^{arctan(\frac 34})$, so the two are the same!

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