Write the expression in the form a+bi , where a and b are real numbers. ((9/2)-(?3/2)i)^3
T tayhalo New member Joined Oct 5, 2010 Messages 6 Oct 5, 2010 #1 Write the expression in the form a+bi , where a and b are real numbers. ((9/2)-(?3/2)i)^3
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Oct 5, 2010 #2 tayhalo said: ((9/2)-(?3/2)i)^3 Your typing means this: (92 − 32⋅i)3\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right)^{3}(29−23⋅i)3 Is this what you intend ? Click to expand... Can you determine the following ? i^3 = (sqrt[3]/2)^3 =
tayhalo said: ((9/2)-(?3/2)i)^3 Your typing means this: (92 − 32⋅i)3\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right)^{3}(29−23⋅i)3 Is this what you intend ? Click to expand... Can you determine the following ? i^3 = (sqrt[3]/2)^3 =
T tayhalo New member Joined Oct 5, 2010 Messages 6 Oct 5, 2010 #3 No the parentheses are around the entire problem
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Oct 5, 2010 #4 Please excuse me. I misread the grouping symbols. (We do not need parentheses around 9/2) (9/2 - sqrt(3)/2 i)^3 So, let's start with the following. (9/2 - sqrt(3)/2 i)^2 What do you get, when you apply the FOIL algorithm to this square ?
Please excuse me. I misread the grouping symbols. (We do not need parentheses around 9/2) (9/2 - sqrt(3)/2 i)^3 So, let's start with the following. (9/2 - sqrt(3)/2 i)^2 What do you get, when you apply the FOIL algorithm to this square ?
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Oct 5, 2010 #6 tayhalo said: 81/4 - 9sqrt(3)/4 - 9sqrt(3)/4 + 3/4 What happened to the factors of i ? The rest of it is okay. Click to expand... (92)⋅(−32⋅i) = −9 34⋅i\displaystyle \left ( \frac{9}{2} \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; - \frac{9 \ \sqrt{3}}{4} \cdot i(29)⋅(−23⋅i)=−49 3⋅i (−32⋅i)⋅(−32⋅i) = 34⋅i2\displaystyle \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; \frac{3}{4} \cdot i^2(−23⋅i)⋅(−23⋅i)=43⋅i2 Do you know the value of i^2 Your result above can be simplified, after you replace i^2 with its value.
tayhalo said: 81/4 - 9sqrt(3)/4 - 9sqrt(3)/4 + 3/4 What happened to the factors of i ? The rest of it is okay. Click to expand... (92)⋅(−32⋅i) = −9 34⋅i\displaystyle \left ( \frac{9}{2} \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; - \frac{9 \ \sqrt{3}}{4} \cdot i(29)⋅(−23⋅i)=−49 3⋅i (−32⋅i)⋅(−32⋅i) = 34⋅i2\displaystyle \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; \frac{3}{4} \cdot i^2(−23⋅i)⋅(−23⋅i)=43⋅i2 Do you know the value of i^2 Your result above can be simplified, after you replace i^2 with its value.
T tayhalo New member Joined Oct 5, 2010 Messages 6 Oct 5, 2010 #7 I have no idea, and the book has no example problem.
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Oct 5, 2010 #8 Your math text must define the symbol i somewhere. I think you just missed it. i = sqrt(-1) Therefore, i^2 = -1 In other words, replace i^2 with -1 in your foiled result above, and simplify. (Also, don't forget to fix the other missing factor of i that I showed you.) Please post your corrections, and we'll go from there.
Your math text must define the symbol i somewhere. I think you just missed it. i = sqrt(-1) Therefore, i^2 = -1 In other words, replace i^2 with -1 in your foiled result above, and simplify. (Also, don't forget to fix the other missing factor of i that I showed you.) Please post your corrections, and we'll go from there.
mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Oct 5, 2010 #10 Okay, you're getting closer. The Rational number 78/4 can be reduced, yes ? Also, the two terms -9 sqrt(3) i/4 and -9 sqrt(3) i/4 are like-terms. I mean, they're identical, yes ? Combine them. With these two simplications, we end up with the following square. (92 − 32⋅i)2 = 392 − 9 32⋅i\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right )^{2} \;=\; \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i(29−23⋅i)2=239−29 3⋅i The original expression has three factors (a cube), and you've now multiplied two of them together (a square). Therefore, to finish, multiply the squared result by the third factor and you're done. (92 − 32⋅i)(392 − 9 32⋅i)\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right ) \left( \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i \right )(29−23⋅i)(239−29 3⋅i)
Okay, you're getting closer. The Rational number 78/4 can be reduced, yes ? Also, the two terms -9 sqrt(3) i/4 and -9 sqrt(3) i/4 are like-terms. I mean, they're identical, yes ? Combine them. With these two simplications, we end up with the following square. (92 − 32⋅i)2 = 392 − 9 32⋅i\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right )^{2} \;=\; \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i(29−23⋅i)2=239−29 3⋅i The original expression has three factors (a cube), and you've now multiplied two of them together (a square). Therefore, to finish, multiply the squared result by the third factor and you're done. (92 − 32⋅i)(392 − 9 32⋅i)\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right ) \left( \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i \right )(29−23⋅i)(239−29 3⋅i)