If α denotes the number of solutions of |1-i|ˣ = 2ˣ and β = (|z|/arg(z)), where z = π/4(1+i)⁴(1-√ π

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If α denotes the number of solutions of |1-i|^x=2^x and β=((|z|)/(arg⁡(z))), where z=π/4(1+i)^4 If α denotes the number of solutions of |1-i|^x=2^x and β=((|z|)/(arg⁡(z))), where z=π/4(1+i)^4
e^(iπ) in 3.14 minutes, using dynamics | DE5 e^(iπ) in 3.14 minutes, using dynamics | DE5
Arg(z − 2 − 4i) = (π / 4)...........#africastruggle. Arg(z − 2 − 4i) = (π / 4)...........#africastruggle.
Arg(((z − 1) / (z − i))) = (π / 4)..........#africastruggle. Arg(((z − 1) / (z − i))) = (π / 4)..........#africastruggle.
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Logarithmic Form to Exponential Form 📚 #Shorts #algebra #math #maths #mathematics #education Logarithmic Form to Exponential Form 📚 #Shorts #algebra #math #maths #mathematics #education
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