edr1c
find all solutions for cos z = 2, z=x+iy?



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silvanx
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2iy = x

silvanx
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2x/i = y

silvanx
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2x/y = i

anonymous
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you need to start with
\[\frac{1}{2}(e^{iz}+e^{iz})=2\]

anonymous
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then to
\[e^{iz}+e^{iz}=4\] and then
\[e^{2iz}+4e^{iz}+1=0\] solve the quadratic equation in \(e^{iz}\)

anonymous
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typo i mean
\[e^{2iz}4e^{iz}+1=0\]

edr1c
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ohh. ic. thx.

anonymous
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yw

Aylen333
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cos never be 2 in maths

Aylen333
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1<cosx<1