1. ## Complex no clarifications

Hi I have a few more doubts!

How do I get from -(±(3-2i))^2 to (3i+2)^2?

I know the - sign in front is i^2, so I multiply it in..but if I multiply i^2 in isn't it (3i^2-2i^3)^2 instead, thus (-3+2i)^2?

Also, is |a-bi| = a+bi?

Thank you so so much!

2. ## Re: Complex no clarifications

\displaystyle \displaystyle \begin{align*} - \left[ \pm \left( 3 - 2i \right) \right]^2 &= -\left( 3 - 2i \right)^2 \\ &= - \left( 9 - 12i - 4 \right) \\ &= - \left( 5 - 12i \right) \\ &= -5 + 12i \end{align*}

and

\displaystyle \displaystyle \begin{align*} \left( 3i + 2 \right)^2 &= -9 + 12i + 4 \\ &= -5 + 12i \\ &= - \left[ \pm \left( 3 - 2i \right) \right]^2 \end{align*}

3. ## Re: Complex no clarifications

Originally Posted by Tutu
Hi I have a few more doubts!

How do I get from -(±(3-2i))^2 to (3i+2)^2?

I know the - sign in front is i^2, so I multiply it in..but if I multiply i^2 in isn't it (3i^2-2i^3)^2 instead, thus (-3+2i)^2?

Also, is |a-bi| = a+bi?

Thank you so so much!
Surely you know that | | is a non-negative real number? So "|a- bi|= a+ bi" is automatically impossible. It possible that you meant to ask if |a- bi|= |a+ bi|. The answer to that is certainly yes: $\displaystyle |a- bi|= \sqrt{a^2+ (-b)^2}= \sqrt{a^2+ b^2}= |a+ bi|$.