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Section 6.6
Can you do 91 and 93 as a guide for me to do a few more? Thank you.
Can you do 91 and 93 as a guide for me to do a few more? Thank you.
Complex Number Equations
- Thread starter nycmathguy
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91
\(x^4+16i=0\)
\(x^4=-16i\)
solutions are:
\(x = 2cos(π/8) - 2i *sin(π/8)\)
\(x = 2cos(-π/8) + i *sin(-π/8)\)
93
\(x^3-(1-i)=0\)
\(x^3=(1-i)\)
\(x^4+16i=0\)
\(x^4=-16i\)
solutions are:
\(x = 2cos(π/8) - 2i *sin(π/8)\)
\(x = 2cos(-π/8) + i *sin(-π/8)\)
93
\(x^3-(1-i)=0\)
\(x^3=(1-i)\)
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91
\(x^4+16i=0\)
\(x^4=-16i\)
Very cool. Maybe I will try a few when time allows.
Feeling discouraged. Went through this section TWICE and just not picking up the material. Bad sign for calculus.
How did you get from x = 4throot {-16i} to x = 2•4throor{-i}?
How did you get from x = cuberoor{1 - i} to the final answer for x?
solutions are:
\(x = 2cos(π/8) - 2i *sin(π/8)\)
\(x = 2cos(-π/8) + i *sin(-π/8)\)
93
\(x^3-(1-i)=0\)
\(x^3=(1-i)\)
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91
\(x^4+16i=0\)
\(x^4=-16i\)
solutions are:
\(x = 2cos(π/8) - 2i *sin(π/8)\)
\(x = 2cos(-π/8) + i *sin(-π/8)\)
93
\(x^3-(1-i)=0\)
\(x^3=(1-i)\)
For 92 and 94, I need a step by step solution starting where I got stuck. Thank you. This completes Chapter 6 for us. I was going to skip Chapters 7 through 9 but decided that this is a bad idea. Chapter 7 begins next Friday. Back to work tomorrow. My weekend is over.
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88.
from here
90.
that is real solution, but we also have two complex solutions
92.
x^6+64*i=0
x^6= -64*i
For z^n=a the solutions are :
k=0,1,..... ,n-1
for n=6, a=-64i
substituting values for a, n, and k you will get following complex solutions:
94. doing this one using same method as in 92, solutions will be
from here
90.
that is real solution, but we also have two complex solutions
92.
x^6+64*i=0
x^6= -64*i
For z^n=a the solutions are :
k=0,1,..... ,n-1
for n=6, a=-64i
substituting values for a, n, and k you will get following complex solutions:
94. doing this one using same method as in 92, solutions will be
- Joined
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- Messages
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- Reaction score
- 422
88.
from here
90.
that is real solution, but we also have two complex solutions
92.
x^6+64*i=0
x^6= -64*i
For z^n=a the solutions are :
k=0,1,..... ,n-1
for n=6, a=-64i
substituting values for a, n, and k you will get following complex solutions:
94. doing this one using same method as in 92, solutions will be
I don't understand this complex equation stuff. I never studied this material back in my college days. Looks terribly advanced for precalculus students. What is the name of the course that goes into great detail concerning such equations? I took precalculus back in 1993. The professor did not teach this stuff.
