In (1+i√2)46 , we can apply exponent rule, (ab)c=acbc . If we do it here, we will get: (1+i)46223 .
Find (1+i√2)46 .
Let's convert 1 + i to polar form so we can use de Moivre's Theorem.
1+i=√2(cosπ4+isinπ4)
Now lets raise both sides of that equation to the power of 46 .
(1+i)46=[√2(cosπ4+isinπ4)]46 (1+i)46=(√2)46(cosπ4+isinπ4)46
And divide both sides by (√2)46
(1+i)46(√2)46=(cosπ4+isinπ4)46 (1+i√2)46=(cosπ4+isinπ4)46
De Moivre's Theorem says....
(cos(x)+isin(x))n=cos(nx)+isin(nx) So....
(cosπ4+isinπ4)46=cos(46π4)+isin(46π4) (1+i√2)46=cos(46π4)+isin(46π4) (1+i√2)46=cos(23π2)+isin(23π2)
Now let's find a reference angle that is coterminal with 23π2 by subtracting 2π five times.
23π2−2π−2π−2π−2π−2π=23π2−10π=23π2−20π2=3π2
So....
(1+i√2)46=cos(3π2)+isin(3π2) (1+i√2)46=0+i(−1) (1+i√2)46=−i