Question

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

Answer 1

Solution:

(a):

$z^4=24i-7$

Put $z=x+iy$

$\Rightarrow (x+iy)^4=24i-7 \\ \Rightarrow [(x+iy)^2]^2=24i-7 \\ \Rightarrow [x^2+i^2y^2+2ixy]^2=24i-7 \\ \Rightarrow [(x^2-y^2)+2ixy]^2=24i-7 \\ \Rightarrow [(x^2-y^2)^2+(2ixy)^2+2(x^2-y^2)(2ixy)]=24i-7 \\ \Rightarrow (x^4+y^4-2x^2y^2)-4x^2y^2+4ix^3y-4ixy^3=24i-7 \\ \Rightarrow x^4+y^4-6x^2y^2+i(4x^3y-4xy^3)=24i-7$

On comparing both sides,

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

Put (ii) in (i),

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

Put $x^2y^2=t$

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

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

Rejecting negative value as it is impossible for real numbers.

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

Put (iii) in (ii)

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

Put this in (iii),

$x=\pm\dfrac{1}{2},\pm2$

Thus, solution is $z=\pm\dfrac12\pm4i;z=\pm2\pm1i$

(b):

$z^ 2 + z −2 = i$

Put $z=x+iy$

$\Rightarrow (x+iy)^ 2 + (x+iy) −2 = i \\ \Rightarrow x^2-y^2+2ixy+x+iy-2=i \\ \Rightarrow (x^2-y^2+x-2)+i(2xy)=0+i$

On comparing,

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

Put (ii) in (i),

$\\ \Rightarrow x^2-(\dfrac{1}{2x})^2+x-2=0 \\ \Rightarrow 4x^4-1+4x^3-8x^2=0 \\ \Rightarrow 4x^4+4x^3-8x^2-1=0 \\ \Rightarrow x\approx \:1.07 ,\:x\approx \:-2.02$

Put these in (ii)

$y=0.47,y=-0.25$

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

(c):

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

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

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

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

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

$z=-\dfrac12+i\cot(\dfrac{\pi}{6});-\dfrac12+i\cot(\dfrac{\pi}{3});-\dfrac12+i\cot(\dfrac{\pi}{2}) \\;-\dfrac12+i\cot(\dfrac{2\pi}{3});-\dfrac12+i\cot(\dfrac{5\pi}{6});-\dfrac12+i\cot(\pi)$

$z=-\dfrac12+i\sqrt3;-\dfrac12+i(\dfrac{1}{\sqrt3});-\dfrac12+i(0) \\;-\dfrac12+i(-\dfrac{1}{\sqrt3});-\dfrac12+i(-\sqrt3);-\dfrac12+i(\infty)$

Thus, solutions are:

$z=-\dfrac12+i\sqrt3;-\dfrac12+i(\dfrac{1}{\sqrt3});-\dfrac12 \\;-\dfrac12-i(\dfrac{1}{\sqrt3});-\dfrac12-i(\sqrt3);\text{not defined}$