√の有理化と同じ要領です
i/5+i=i(5-i)/(5+i)(5-i)=(5i+1)/25+1=(1+5i)/26
(1-2i)/(1+2i)=(1-2i)(1-2i)/(1+2i)(1-2i)
=(1-4i-4)/(1+4)=-(3+4i)/5
√の有理化と同じ要領です
i/5+i=i(5-i)/(5+i)(5-i)=(5i+1)/25+1=(1+5i)/26
(1-2i)/(1+2i)=(1-2i)(1-2i)/(1+2i)(1-2i)
=(1-4i-4)/(1+4)=-(3+4i)/5
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