Question

23
-
evaluate j^23 in complex numbers​

Answer 1

Answer:

that very easy lesson of complex numbers

Step-by-step explanation:

first u remember the simple rules

$j = \sqrt{-1} \\j^{2} = -1\\j^{3} = j^{2} \times j = -1 \times j = -j\\j^{4} = -1 \times -1 = 1$

given is

$j^{23} = j^{20} \times j^{3}$

you see j power of 20. you know 20 can divide by 4 therefore,

$(j^{4})^{5} \times j^{3} = 1^5 \times j^3 [ \because j^4 = 1] \\j^3= -j [\because j^3 = j^2 \times j = -1 \times j = -j]\\\therefore j^{23} = -j$