# Math Help - Complex numbers in this form: re^(a+bi) to polar form (r,thea)

1. ## Complex numbers, what are the values of r and theta?

Write in the polar form ,.
a)

I know that I should use this: (e^a)(cosb+isinb)
Which is: (e^4)(cos1+isin1)
But then what? How do I get 7e^(a+i) into a+bi form?

*To be more precise what are r=? and theta=? (without decimals)

b)

For question b r=

But does Theta equal? I got but it's wrong.

If there are other posts that could help please feel free to send me links.

2. ## Re: Complex numbers in this form: re^(a+bi) to polar form (r,thea)

$e^{a+ bi}= e^a e^{bi}= e^a(cos(b+ i sin b)$ which is already a "polar form" with $r= e^a$ and $\theta= b$.
$7e^{a+i}= 7e^a(e^{i})= 7e^a(cos(1)+ i sin(1)$

$\frac{2i}{7e^{4+i}}= \frac{2}{7}(i)e^{-4- i}= \frac{2}{7}(e^{-\pi/4})(e^{-4}(e^{-i})= \frac{2}{7}e^{-4}e^{-(1+ \pi/4)i}$

3. ## Re: Complex numbers in this form: re^(a+bi) to polar form (r,thea)

Thank you but this looks like Chinese to me. I basically need the value of r and theta that's all.
If you're saying that r= and theta=
r is correct but theta is wrong

4. ## Re: Complex numbers, what are the values of r and theta?

Originally Posted by SillyMath
b)

For question b r=

But does Theta equal? I got but it's wrong.
In standard form, this complex number should be
$\frac{-\sqrt{3}-1}{2} + \frac{\sqrt{3}-1}{2}i$

Which means that theta should be
$\tan^{-1} \left( \frac{\sqrt{3} - 1}{-\sqrt{3} - 1}\right) + \pi$

Note: I normally use the
$\tan \theta = \frac{b}{a}$
formula to find theta instead of inverse sine.

5. ## Re: Complex numbers, what are the values of r and theta?

Thank you to all, this really helped me
oh and the answer for a) theta=(pi/2)-1