2016年2月10日 · This is the same as i The complex i repeats after 4 powers This can be seen with i^1 = i i^2 = sqrt(-1)^2 = -1 i^3 = i^2 * i = -1 * i => -i i^4 = i^2 * i^2 => -1 * -1 >>> 1 After this, we can take the mod 4 of the exponent (in this case we can just do the next exponent, try it out and of course the answer will be the same as i ^ 1 ) Thus, the answer is i.

2016年2月2日 · 12/13 - 5/13 i >To divide 2 complex numbers , multiply the numerator and denominator by the' complex conjugate' of the denominator. If a + bi is a complex number thencolor(red)(" a - bi is the conjugate") This ensures that the denominator is real , as shown below. multiplying a complex number and it's conjugate. (a + bi )(a - bi ) = a^2 + abi - abi - bi^2 = a^2 + b^2 which is real . [ remember ...

2016年8月23日 · To use DeMoivre's Theorem, we need to convert . #z=-sqrt3+i=x+iy# into Polar Form, i.e., #r(costheta+isintheta), where, r>0, &, theta in (-pi,pi]#.

2016年3月16日 · (1-i)^5 = -4+4i The exponent in this case isn't too difficult to calculate directly using the binomial theorem

2016年12月3日 · The answer is =55.9(-0.68+0.73i) To use DeMoivre's theorem, we need to change to the trigonometric form of the complex numbers. (costheta+isintheta)^n=cosntheta+isinntheta Here, z=2+i ∥z∥=sqrt(4+1)=sqrt5 z=sqrt5(2/sqrt5+i/sqrt5) z=sqrt5(costheta+isintheta) Therefore, costheta=2/sqrt5 sintheta=1/sqrt5 theta=0.46 rd z=sqrt5(cos0.46+isin0.46) Therefore, z^5=(sqrt5(cos0.46+isin0.46))^5 =(sqrt5 ...

2017年1月5日 · How do you simplify #(5-i)/(5+i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex ...

2015年4月26日 · How do I use DeMoivre's theorem to find #(1+i)^5#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers. 1 Answer

2015年12月22日 · Multiply the numerator and denominator by the conjugate of the denominator to find (2-i)/(5+i)= 9/26 - 7/26i The conjugate of a complex number a+bi is a-bi. The product of a complex number and its conjugate is a real number. We will use that fact here.

2016年10月22日 · How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#?

2015年4月15日 · Viewed as a right angled triangle tan(x)=5/12 can be thought of as the ratio of opposite to adjacent sides in a triangle with sides 5, 12 and 13 (where 13 is derived from the Pythagorean Theorem) So sin(x) = 5/13 and cos(x) = 12/13