Complex number question

[Jared123]

When I attempted it my answer was -9sqrt(2) - 9sqrt(2) and 9sqrt(2) + 9sqrt(2) which I know is wrong

 

[Romsek]

[MATH]z^2 = 36i = 36 e^{i \pi/2}[/MATH]
[MATH]z^n = r e^{i\theta} \Rightarrow z = r^{1/n} e^{i (\theta+2\pi k)/n},~k = 0, 1, \dots, n-1[/MATH]
Can you solve for $\displaystyle z$ now?

 

[Jared123]

[MATH]z^2 = 36i = 36 e^{i \pi/2}[/MATH]
[MATH]z^n = r e^{i\theta} \Rightarrow z = r^{1/n} e^{i (\theta+2\pi k)/n},~k = 0, 1, \dots, n-1[/MATH]
Can you solve for $\displaystyle z$ now?

thank you this helped. Im not sure if its correct but the andwer im getting now is 3 sqrt(2) + 3 sqrt(2) i and -3 sqrt(2)- 3 sqrt(2) i

 

[Dr.Peterson]

If you need additional help, please show your work, not just your answer, so we can help you identify specific errors. You may have used a different, but valid, method, and just made a silly mistake (e.g. dividing 36 by 2 instead of taking its square root).

 

[Jared123]

thank you this helped. Im not sure if its correct but the andwer im getting now is 3 sqrt(2) + 3 sqrt(2) i and -3 sqrt(2)- 3 sqrt(2) i

sorry about that. Here is my working out

 

[pka]

sorry about that. Here is my working out
View attachment 17747

Both of the above answers are correct. But you forgot to put an $i$ in the last line in the right.
Your notation leaves much to be desired. I would write it as $\pm(3\sqrt2+3\sqrt2\bf{i})$

 

[topsquark]

$\pm(3\sqrt2+3\sqrt2\bf{i})$

Frankly I'd write it as $$\pm (3 \sqrt{2} + 3i \sqrt{2} )$$ to make sure it's clear that the i isn't inside the square root. But that might be a Physicist thing.

-Dan

 

[Deleted member 4993]

Frankly I'd write it as $$\pm (3 \sqrt{2} + 3i \sqrt{2} )$$ to make sure it's clear that the i isn't inside the square root. But that might be a Physicist thing.

-Dan

That is being kind others thing...

 

[pka]

Frankly I'd write it as $$\pm (3 \sqrt{2} + 3i \sqrt{2} )$$ to make sure it's clear that the i isn't inside the square root. But that might be a Physicist thing.

It may be a mathematical thing, we a accustomed to say a complex number is written as $a+b\bf{i}$ where both $a~\&~b$ are real numbers.
Hence $3\sqrt2+3\sqrt2\bf{i}$.