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Question Number 9923 by konen last updated on 16/Jan/17

i=(√(−1)) z= (1−i)^(17) −(1+i)^(17) ⇒ z=?

$$\mathrm{i}=\sqrt{−\mathrm{1}} \\ $$$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \:\Rightarrow\:\mathrm{z}=?\: \\ $$

Commented by RasheedSoomro last updated on 16/Jan/17

z= (1−i)^(17) −(1+i)^(17) 1−i=(1−i)×(i/i)=((i−i^2 )/i)=((1+i)/i) z =(((1+i)/i))^(17) −(1+i)^(17) =(((1+i)^(17) )/(i^(16) .i))−(1+i)^(17) =(((1+i)^(17) −i(1+i)^(17) )/i) =(((1+i)^(17) (1−i))/i) =(1+i)^(17) ×((1−i)/i)×(i/i) =(1+i)^(17) ×−(1+i) =−(1+i)^(18) =−{(1+i)^2 }^9 (1+i)^2 =1+2i+i^2 =2i =−(2i)^9 =−2^9 .(i^4 )^2 .i =−512 i ∵ i^4 =1

$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \\ $$$$\mathrm{1}−\mathrm{i}=\left(\mathrm{1}−\mathrm{i}\right)×\frac{\mathrm{i}}{\mathrm{i}}=\frac{\mathrm{i}−\mathrm{i}^{\mathrm{2}} }{\mathrm{i}}=\frac{\mathrm{1}+\mathrm{i}}{\mathrm{i}} \\ $$$$\:\mathrm{z}\:=\left(\frac{\mathrm{1}+\mathrm{i}}{\mathrm{i}}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \\ $$$$\:\:\:=\frac{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} }{\mathrm{i}^{\mathrm{16}} .\mathrm{i}}−\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \\ $$$$\:\:\:=\frac{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} −\mathrm{i}\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} }{\mathrm{i}} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{i}} \\ $$$$\:\:\:\:=\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} ×\frac{\mathrm{1}−\mathrm{i}}{\mathrm{i}}×\frac{\mathrm{i}}{\mathrm{i}} \\ $$$$\:\:\:\:\:=\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} ×−\left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\:\:\:\:\:=−\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{18}} \\ $$$$\:\:\:\:\:=−\left\{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} \right\}^{\mathrm{9}} \:\:\: \\ $$$$\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2i}+\mathrm{i}^{\mathrm{2}} =\mathrm{2i} \\ $$$$\:\:\:\:\:=−\left(\mathrm{2i}\right)^{\mathrm{9}} \\ $$$$\:\:\:\:\:=−\mathrm{2}^{\mathrm{9}} .\left(\mathrm{i}^{\mathrm{4}} \right)^{\mathrm{2}} .\mathrm{i} \\ $$$$\:\:\:\:\:\:=−\mathrm{512}\:\mathrm{i}\:\:\:\:\because\:\mathrm{i}^{\mathrm{4}} =\mathrm{1} \\ $$

Answered by RasheedSoomro last updated on 16/Jan/17

z= (1−i)^(17) −(1+i)^(17) ⇒ z=? z= (1−i)^(17) −(1+i)^(17) ={(1−i)^2 }^8 (1−i)−{(1+i)^2 }^8 (1+i) (1−i)^2 =1−2i+i^2 =−2i (1+i)^2 =1+2i+i^2 =2i =(−2i)^8 (1−i)−(2i)^8 (1+i) =(2i)^8 (1−i)−(2i)^8 (1+i) =(2i)^8 {1−i−1−i} =(2i)^8 (−2i) =−2^9 .i^8 .i^ =−512.(i^4 )^2 .i =−512 i ∵ i^4 =1

$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \:\Rightarrow\:\mathrm{z}=?\: \\ $$$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \\ $$$$\:\:\:=\left\{\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{2}} \right\}^{\mathrm{8}} \left(\mathrm{1}−\mathrm{i}\right)−\left\{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} \right\}^{\mathrm{8}} \left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{2i}+\mathrm{i}^{\mathrm{2}} =−\mathrm{2i} \\ $$$$\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2i}+\mathrm{i}^{\mathrm{2}} =\mathrm{2i} \\ $$$$\:\:\:\:\:=\left(−\mathrm{2i}\right)^{\mathrm{8}} \left(\mathrm{1}−\mathrm{i}\right)−\left(\mathrm{2i}\right)^{\mathrm{8}} \left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2i}\right)^{\mathrm{8}} \left(\mathrm{1}−\mathrm{i}\right)−\left(\mathrm{2i}\right)^{\mathrm{8}} \left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2i}\right)^{\mathrm{8}} \left\{\mathrm{1}−\mathrm{i}−\mathrm{1}−\mathrm{i}\right\} \\ $$$$\:\:\:\:\:\:=\left(\mathrm{2i}\right)^{\mathrm{8}} \left(−\mathrm{2i}\right) \\ $$$$\:\:\:\:\:\:=−\mathrm{2}^{\mathrm{9}} .\mathrm{i}^{\mathrm{8}} .\mathrm{i}^{} \\ $$$$\:\:\:\:\:\:=−\mathrm{512}.\left(\mathrm{i}^{\mathrm{4}} \right)^{\mathrm{2}} .\mathrm{i} \\ $$$$\:\:\:\:\:\:=−\mathrm{512}\:\mathrm{i}\:\:\:\:\because\:\:\mathrm{i}^{\mathrm{4}} =\mathrm{1} \\ $$

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