# Complex no: Where did I go wrong

• November 16th 2012, 11:57 PM
Tutu
Complex no: Where did I go wrong
Another one! I'm really sorry. This time I'm not sure where I went wrong.

Express in polar form, (1/(sqrt(21) - sqrt(7)i)*)
Conjugate, so I got (1/(sqrt(21) + sqrt(7)i))
Let z = sqrt(21) + sqrt(7)i
In modulus argument form ie. cis form, I got sqrt(30)cis(pi/6)
For the modulus of (1/(sqrt(21) - sqrt(7)i)*),
I got 1/sqrt(30)

Thank you very much for your time,
J.
• November 17th 2012, 12:04 AM
Prove It
Re: Complex no: Where did I go wrong
Quote:

Originally Posted by Tutu
Another one! I'm really sorry. This time I'm not sure where I went wrong.

Express in polar form, (1/(sqrt(21) - sqrt(7)i)*)
Conjugate, so I got (1/(sqrt(21) + sqrt(7)i))
Let z = sqrt(21) + sqrt(7)i
In modulus argument form ie. cis form, I got sqrt(30)cis(pi/6)
For the modulus of (1/(sqrt(21) - sqrt(7)i)*),
I got 1/sqrt(30)
\displaystyle \begin{align*} \frac{1}{\sqrt{21} - i\sqrt{7}} &= \frac{\sqrt{21} + i\sqrt{7}}{\left( \sqrt{21} - i\sqrt{7} \right) \left( \sqrt{21} + i\sqrt{7} \right)} \\ &= \frac{\sqrt{21} + i\sqrt{7}}{\left( \sqrt{27} \right)^2 + \left( \sqrt{7} \right)^2 } \\ &= \frac{\sqrt{21} + i\sqrt{7}}{34} \\ &= \frac{\sqrt{27}}{34} + i\frac{\sqrt{7}}{34} \end{align*}
Now that it's in \displaystyle \begin{align*} a + i\,b \end{align*} form, try converting to polars.