Coursera: analysis complex kind Week 3.3. Complex exponential function.

Jul 18, 2016

Background:

e^{i\theta} = cos(\theta) + i sin (\theta)

Hence, e^{i(\theta + 2k \pi)} = cos(\theta + 2k \pi) + i sin (\theta + 2k \pi) =  cos(\theta) + i sin (\theta) = e^{i\theta}

Overall:  e^{i(\theta + 2k \pi)} = e^{i\theta}

Definition

f(z) = e^x(cos y + i sin y) = e^{(x + iy)}

When y = 0, this is same as real exponential function

Properties

| e^z | = |e^x | |e^{iy} | = e^x (since e^{i\theta} sits on the unit circle, thus has size = 1)

Also, arg(e^z) = arg(e^{x+iy}) =  arg(e^{x}.e^{iy}) = arg(e^{x}) + arg(e^{iy})

from above, e^x is the modulus or size of radius of e^z, then (e^{x}.e^{iy})  is in polar form r.e^{i\theta} and hence y is argument of e^{(x + iy)}

Note: arg(z_1z_2) = arg(z_1) + arg(z_2) (from earlier lecture – ref required) is not used in determining arg(e^{x+iy}) and do not confuse this with arg({x+iy})

e^{z + 2{\pi}i} = e^{z}

e^{z + w} = e^{z}.e^{w} by setting z = x+iy and w= u + iv

\frac{1}{e^z} = e^{-z } , since {e^z}.e^{-z } = e^0 = 1

e^z = 1 \iff e^{x+iy} = e^x.e^{iy} = 1

\implies modulus (length) = 1 so x =0, and argument = 2k \pi

e^z = 1 \iff z = 2ik \pi

Hence , when e^z = e^w then e^{z - w} = 1 therefore z – w = 2ik \pi

So z  = w + 2ik \pi results in e^z = e^w

<< to be done : mapping w = e^z >>

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One Response to “Coursera: analysis complex kind Week 3.3. Complex exponential function.”


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