- #1
suspenc3
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2 Questions:
(1.)Carry out the indicated calculation:
[tex](\frac{-6+2i}{1-8i})^2[/tex]
=[tex]\frac{36-24i+4i^2}{1-16i+64i^2}[/tex]
since [tex]i^2=-1[/tex]
=[tex]\frac{32-24i}{-63-16i}[\frac{-63+16i}{-63+16i}][/tex]
I carry out the math and get an answer of:
[tex]\frac{-2400+1512i}{4225}[/tex]
I must be doing something wrong with the conjugate, as far as i can tell it looks right, but for some reason my signs should be switched to give me the correct answer of:
[tex]\frac{-1632+2024i}{4225}[/tex]
(2.)determine the set of all z satisfying the given equation or inequality:
[tex]|z-2i|<=|z+1+i| & |z|> 4[/tex]
I solved this one down to:
[tex]x^2+y^2-4y+4<=x^2+y^2+2x+2y+2[/tex] & [tex]x^2+y^2>4[/tex]
what do I do to simplify it?
Thanks.
(1.)Carry out the indicated calculation:
[tex](\frac{-6+2i}{1-8i})^2[/tex]
=[tex]\frac{36-24i+4i^2}{1-16i+64i^2}[/tex]
since [tex]i^2=-1[/tex]
=[tex]\frac{32-24i}{-63-16i}[\frac{-63+16i}{-63+16i}][/tex]
I carry out the math and get an answer of:
[tex]\frac{-2400+1512i}{4225}[/tex]
I must be doing something wrong with the conjugate, as far as i can tell it looks right, but for some reason my signs should be switched to give me the correct answer of:
[tex]\frac{-1632+2024i}{4225}[/tex]
(2.)determine the set of all z satisfying the given equation or inequality:
[tex]|z-2i|<=|z+1+i| & |z|> 4[/tex]
I solved this one down to:
[tex]x^2+y^2-4y+4<=x^2+y^2+2x+2y+2[/tex] & [tex]x^2+y^2>4[/tex]
what do I do to simplify it?
Thanks.