What is (e^z)\circ (z^4+iz^2+(3-i)) ?

The solution to (e^z)\circ (z^4+iz^2+(3-i)) is (\circ e^zz^4+3\circ e^z)+i(-\circ e^z+\circ e^zz^2)

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simplify (e^z)\circ (z^4+iz^2+(3-i))

simplify

(e^{z}) \circ (z^{4} + i z^{2} + (3 - i))

(\circ e^{z} z^{4} + 3 \circ e^{z}) + i (- \circ e^{z} + \circ e^{z} z^{2})

Solution steps

(e^{z}) \circ (z^{4} + i z^{2} + (3 - i))

Remove parentheses:

(a) = a

= e^{z} \circ (z^{4} + i z^{2} + 3 - i)

Rewrite

e^{z} \circ (z^{4} + i z^{2} + 3 - i)

in standard complex form:

(\circ e^{z} z^{4} + 3 \circ e^{z}) + (\circ e^{z} z^{2} - \circ e^{z}) i

= (\circ e^{z} z^{4} + 3 \circ e^{z}) + (\circ e^{z} z^{2} - \circ e^{z}) i

\frac{2 ( 33 - 417 )}{3}

\frac{x ^{5} + 34.5 x ^{4} + 64 x ^{3} + 54}{x ^{4} ( x - 1.5 ) ^{2}}

\frac{( a - b )}{a b - b a}

\frac{x ^{2}}{( 8 + x ) ^{3}}

(4 i - 2 k) l e f t (i + j + k)

What is (e^z)\circ (z^4+iz^2+(3-i)) ?
The solution to (e^z)\circ (z^4+iz^2+(3-i)) is (\circ e^zz^4+3\circ e^z)+i(-\circ e^z+\circ e^zz^2)