Question #221374

Solve the equation (a) z 4 = 24i − 7, (b) z 2 + z −2 = i, (c) z 6 = (1 + z) 6 


1
Expert's answer
2021-08-03T10:16:07-0400

Solution:

(a):

z4=24i7z^4=24i-7

Put z=x+iyz=x+iy

(x+iy)4=24i7[(x+iy)2]2=24i7[x2+i2y2+2ixy]2=24i7[(x2y2)+2ixy]2=24i7[(x2y2)2+(2ixy)2+2(x2y2)(2ixy)]=24i7(x4+y42x2y2)4x2y2+4ix3y4ixy3=24i7x4+y46x2y2+i(4x3y4xy3)=24i7\Rightarrow (x+iy)^4=24i-7 \\ \Rightarrow [(x+iy)^2]^2=24i-7 \\ \Rightarrow [x^2+i^2y^2+2ixy]^2=24i-7 \\ \Rightarrow [(x^2-y^2)+2ixy]^2=24i-7 \\ \Rightarrow [(x^2-y^2)^2+(2ixy)^2+2(x^2-y^2)(2ixy)]=24i-7 \\ \Rightarrow (x^4+y^4-2x^2y^2)-4x^2y^2+4ix^3y-4ixy^3=24i-7 \\ \Rightarrow x^4+y^4-6x^2y^2+i(4x^3y-4xy^3)=24i-7

On comparing both sides,

x4+y46x2y2=7, 4x3y4xy3=24(x2y2)2(2xy)2=7, 4xy(x2y2)=24(x2y2)2(2xy)2=7 ...(i), (x2y2)=6xy ...(ii)x^4+y^4-6x^2y^2=-7,\ 4x^3y-4xy^3=24 \\\Rightarrow (x^2-y^2)^2-(2xy)^2=-7,\ 4xy(x^2-y^2)=24 \\\Rightarrow (x^2-y^2)^2-(2xy)^2=-7\ ...(i),\ (x^2-y^2)=\dfrac{6}{xy}\ ...(ii)

Put (ii) in (i),

(6xy)2(2xy)2=736x2y24x2y2=7\\\Rightarrow (\dfrac6{xy})^2-(2xy)^2=-7 \\\Rightarrow \dfrac{36}{x^2y^2}-4x^2y^2=-7

Put x2y2=tx^2y^2=t

36t4t=7\\\Rightarrow \dfrac{36}{t}-4t=-7

t=94,t=4x2y2=94,x2y2=4\Rightarrow t=-\dfrac{9}{4},\:t=4 \\ \Rightarrow x^2y^2=-\dfrac{9}{4},\:x^2y^2=4

Rejecting negative value as it is impossible for real numbers.

xy=±2x=±2y ...(iii)\Rightarrow xy=\pm2 \\ \Rightarrow x=\dfrac{\pm2}{y} \ ...(iii)

Put (iii) in (ii)

(4y2y2)=6±2=±34y2y2=3;4y2y2=3y=4,1;y=1,4y=±4,±1(\dfrac{4}{y^2}-y^2)=\dfrac{6}{\pm2}=\pm3 \\\Rightarrow \dfrac{4}{y^2}-y^2=3; \dfrac{4}{y^2}-y^2=-3 \\ \Rightarrow y=-4,1;y=-1,4 \\ \Rightarrow y=\pm4,\pm1

Put this in (iii),

x=±12,±2x=\pm\dfrac{1}{2},\pm2

Thus, solution is z=±12±4i;z=±2±1iz=\pm\dfrac12\pm4i;z=\pm2\pm1i

(b):

z2+z2=iz^ 2 + z −2 = i

Put z=x+iyz=x+iy

(x+iy)2+(x+iy)2=ix2y2+2ixy+x+iy2=i(x2y2+x2)+i(2xy)=0+i\Rightarrow (x+iy)^ 2 + (x+iy) −2 = i \\ \Rightarrow x^2-y^2+2ixy+x+iy-2=i \\ \Rightarrow (x^2-y^2+x-2)+i(2xy)=0+i

On comparing,

x2y2+x2=0; 2xy=1x2y2+x2=0 ...(i); y=12x ...(ii)x^2-y^2+x-2=0;\ 2xy=1 \\\Rightarrow x^2-y^2+x-2=0\ ...(i);\ y=\dfrac{1}{2x} \ ...(ii)

Put (ii) in (i),

x2(12x)2+x2=04x41+4x38x2=04x4+4x38x21=0x1.07,x2.02\\ \Rightarrow x^2-(\dfrac{1}{2x})^2+x-2=0 \\ \Rightarrow 4x^4-1+4x^3-8x^2=0 \\ \Rightarrow 4x^4+4x^3-8x^2-1=0 \\ \Rightarrow x\approx \:1.07 ,\:x\approx \:-2.02

Put these in (ii)

y=0.47,y=0.25y=0.47,y=-0.25

Thus, solutions are z=1.07+0.47i; z=2.020.25iz=1.07+0.47i;\ z=-2.02-0.25i

(c):

z6=(1+z)6(z1+z)6=1z1+z=1z1+z=eikπ/3z ^6 = (1 + z)^ 6 \\\Rightarrow (\dfrac{z}{1+z})^6=1 \\\Rightarrow \dfrac{z}{1+z}=1 \\\Rightarrow \dfrac{z}{1+z}=e^{ik\pi/3} , where k=1,2,..,6k=1,2,..,6

z=eikπ/31eikπ/3\\\Rightarrow z=\dfrac{e^{ik\pi/3}}{1-e^{ik\pi/3}}

 Simplify this using trigonometric form and double angle identites to get:

z=12+icot(kπ6)z=-\dfrac12+i\cot(\dfrac{k\pi}{6})

Putting k=1,2,...,6k=1,2,...,6

z=12+icot(π6);12+icot(π3);12+icot(π2);12+icot(2π3);12+icot(5π6);12+icot(π)z=-\dfrac12+i\cot(\dfrac{\pi}{6});-\dfrac12+i\cot(\dfrac{\pi}{3});-\dfrac12+i\cot(\dfrac{\pi}{2}) \\;-\dfrac12+i\cot(\dfrac{2\pi}{3});-\dfrac12+i\cot(\dfrac{5\pi}{6});-\dfrac12+i\cot(\pi)

z=12+i3;12+i(13);12+i(0);12+i(13);12+i(3);12+i()z=-\dfrac12+i\sqrt3;-\dfrac12+i(\dfrac{1}{\sqrt3});-\dfrac12+i(0) \\;-\dfrac12+i(-\dfrac{1}{\sqrt3});-\dfrac12+i(-\sqrt3);-\dfrac12+i(\infty)

Thus, solutions are:

z=12+i3;12+i(13);12;12i(13);12i(3);not definedz=-\dfrac12+i\sqrt3;-\dfrac12+i(\dfrac{1}{\sqrt3});-\dfrac12 \\;-\dfrac12-i(\dfrac{1}{\sqrt3});-\dfrac12-i(\sqrt3);\text{not defined}


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