Finding roots of an exponential equation

Akash47 · 2019-11-01T05:41:34+00:00

I know that both 3^x and e^x can't be 0 for real x. Then x^2-4=0 is the only choice and we get x=2,-2. Am I right? Or should I add something?

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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Thread 'Find the polar form of a complex number'

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