# Complex Numbers Basics!

• December 9th 2012, 07:01 PM
Tutu
Complex Numbers Basics!
Hi I'm not very sure if this is in the correct forum, but I sincerely apologize if it is not.

If a is ±3 and b is ±1,
in forming the complex number a+bi,
is it right to say that the possible combinations include -3-i and 3+i?
I was doing a question on finding the square root of 8-6i, and got down to a i= ±3 and b = ±1, I don't really know which to use! My friend used 3-i to form
±(3-i)^2, but why can't I use ±(3+i)^2 etc?

2.) Solve the simultaneous equations iz+2w=2 and z-(1+i)w=4 and give your answer in the form a+ib where a and b are real.

I tried making w and z the subject respectively, and equated the equations together, but then I realized I was getting a value for z and w, nothing that gives me a+bi..how do I get a and b?

Thank you so much for your time! Really appreciate it!
• December 9th 2012, 07:12 PM
MarkFL
Re: Complex Numbers Basics!
For the first one, you probably found $a=\pm3$ and $b=\mp1$.
• December 9th 2012, 08:57 PM
Prove It
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Hi I'm not very sure if this is in the correct forum, but I sincerely apologize if it is not.

If a is ±3 and b is ±1,
in forming the complex number a+bi,
is it right to say that the possible combinations include -3-i and 3+i?
I was doing a question on finding the square root of 8-6i, and got down to a i= ±3 and b = ±1, I don't really know which to use! My friend used 3-i to form
±(3-i)^2, but why can't I use ±(3+i)^2 etc?

2.) Solve the simultaneous equations iz+2w=2 and z-(1+i)w=4 and give your answer in the form a+ib where a and b are real.

I tried making w and z the subject respectively, and equated the equations together, but then I realized I was getting a value for z and w, nothing that gives me a+bi..how do I get a and b?

Thank you so much for your time! Really appreciate it!

The easiest way to find square roots of complex numbers is to write them in the exponential form first.

\displaystyle \begin{align*} \left( 8 - 6i \right)^{\frac{1}{2}} &= \left[ 2 \left( 4 - 3i \right) \right]^{\frac{1}{2}} \\ &= \sqrt{2} \left( 4 - 3i \right)^{\frac{1}{2}} \\ &= \sqrt{2} \left( 5 \, e^{ -\arctan{ \frac{3}{4} } } \right)^{\frac{1}{2}} \\ &= \sqrt{2}\, \sqrt{5} \, e^{-\frac{1}{2} \arctan{\frac{3}{4}}} \\ &= \sqrt{10} \left[ \cos{\left( -\frac{1}{2} \arctan{\frac{3}{4}} \right)} + i \sin{\left( -\frac{1}{2} \arctan{\frac{3}{4}} \right) } \right] \end{align*}

Now simplify, and remember that the second root will have the same magnitude but separated by an angle of \displaystyle \begin{align*} \pi \end{align*}.