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Mathematics 44 Online
OpenStudy (anonymous):

write the expression as a complex number in standard form. (2+3i)to the 2nd power

OpenStudy (whpalmer4):

When something is to the 2nd power, it means <something>*<something>. Here, we have \((2+3i)\) to the second power, which means \[(2+3i)^2 = (2+3i)(2+3i) \]

OpenStudy (whpalmer4):

Now we need to multiply that out and simplify in order to put it in standard form \((a+bi)\)

OpenStudy (whpalmer4):

\[(2+3i)(2+3i) = 2(2+3i) + 3i(2+3i) \]\[= 2*2 + 2*3i + 3i*2 + 3i*3i\]Can you collect like terms and simplify? Remember that \(i^2=-1\)

OpenStudy (anonymous):

you add the 2 together

OpenStudy (anonymous):

thats not right is it

OpenStudy (whpalmer4):

What is 2*2? What is 2*3i? What is 3i*2? What is 3i*3i? What do you get when you add all of those together?

OpenStudy (anonymous):

4+21i

OpenStudy (whpalmer4):

2*2 = 4 2*3i = 6i 3i*2 = 6i 3i*3i = ?

OpenStudy (anonymous):

10i

OpenStudy (whpalmer4):

\[3i * 3i = 3*i * 3*i = 3*3*i*i = \]

OpenStudy (anonymous):

9 to the 2nd power

OpenStudy (whpalmer4):

No.... \[3*3*i*i = 9*i*i = 9i^2\]But \(i^2 = -1\) so that simplifies further to \[3*3*i*i = 9i^2 = 9(-1) = -9\] So our whole expression is \[4+6i+6i-9=\]

OpenStudy (anonymous):

i want to do a problem i want you to tell me if its right

OpenStudy (whpalmer4):

Okay, right after you finish this one :-)

OpenStudy (anonymous):

the 6 dont cancel out right

OpenStudy (whpalmer4):

Just add up the pieces...4-9=? 6i+6i=?

OpenStudy (anonymous):

13 and 12i

OpenStudy (whpalmer4):

4-9 = 13? Is that your final answer? :-)

OpenStudy (anonymous):

yeah

OpenStudy (whpalmer4):

If 4-9 = 13, what does 4+9 = ?

OpenStudy (anonymous):

its 5

OpenStudy (whpalmer4):

Sigh. |dw:1370894141448:dw| If you start out at 4 on that number line, and you go 9 to the left, where do you end up?

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