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Question Number 161991 by mkam last updated on 25/Dec/21
Commented by mkam last updated on 25/Dec/21
?????
Answered by aleks041103 last updated on 25/Dec/21
z=reitw=1re−it(a)0<r<1⇒1r∈(1,∞)t∈[0,π/2]⇒−t∈[3π2,2π]⇒{z:0<∣z∣<1,0≤argz≤π/2}→w=1z{w:∣w∣>1,3π2≤arg(w)≤2π}⊂C(b)3⩽r⇒1r∈[13,∞)∪{∞}t∈[0,π]⇒−t∈[π,2π]⇒{z:∣z∣⩽3,0≤argz≤π}→{w:∣w∣⩾13,π≤arg(w)≤2π}∪{∞}⊂(C∪{∞})
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