+0  
 
+3
523
2
avatar+1860 

Find all of the fifth roots of the complex number \(4+32i\).  Put your answers in the rectangular form \(a+bi\) and in the polar form \(z=re^(i\Theta )\).  Please show how you got to your answers.

gibsonj338  Nov 24, 2015

Best Answer 

 #2
avatar+1860 
+10

\((4+32i)^(1/5)\)

 

\(r=\sqrt(a^2+b^2)\)

 

\(r=\sqrt(4^2+32^2)\)

 

\(r=\sqrt(16+1024)\)

 

\(r=\sqrt1040\)

 

\(r=4\sqrt65\)

 

\(tan(\Theta)=b/a\)

 

\(tan(\Theta)=32/4\)

 

\(tan(\Theta)=8\)

 

\(\Theta=tan^-1(8)\)

 

\(\Theta ≈1.4464413322481\)

 

 

\(z=r*e^(i*\Theta)\)

 

\(z≈4\sqrt65*e^(i*1.4464413322481)\)

 

\(z^(1/5)≈(4\sqrt65*e^(i*1.4464413322481))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*1.4464413322481*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.0766143512748)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.0766143512748)+i*sin(1.0766143512748))\)

 

\(z≈2.003103242348*(0.4743116564868+i*0.88035700289064)\)

 

\(z≈0.9500952169545+i*1.763445966914\)

 

\(z≈0.9500952169545+1.763445966914i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*7.7296266394277))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*7.7296266394277*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.5459253278855)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.5459253278855)+i*sin(1.5459253278855))\)

 

\(z≈2.003103242348*(0.024868434927217+i*0.99969073264899)\)

 

\(z≈-0.049814942634829+i*0.0001545372385216\)

 

\(z≈-0.049814942634829+0.0001545372385216i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*14.012811946607))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*14.012811946607*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.6955283616309)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.6955283616309)+i*sin(1.6955283616309))\)

 

\(z≈2.003103242348*(-0.12440885450163+i*0.99223104009177)\)

 

\(z≈-0.24920377983349+i*1.9875412136019\)

 

\(z≈-0.24920377983349+1.9875412136019i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*20.2959972538))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*20.2959972538*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*4.059994507574)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(4.05994507574)+i*sin(4.05994507574))\)

 

\(z≈2.003103242348*(-0.60772245598411+i*-0.7941494925344)\)

 

\(z≈-1.2173308220295+i*1.5907634234047\)

 

\(z≈-1.2173308220295+1.5907634234047i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*26.5918256098))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*26.5918256098*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*5.315836512196)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(5.315836512196)+i*sin(5.315836512196))\)

 

\(z≈2.003103242348*(0.56748448302716+i*-0.82338409112843)\)

 

\(z≈-1.1367300079339+i*-1.6493233426371\)

 

\(z≈-1.1367300079339+-1.6493233426371i\)

 

\(z≈-1.1367300079339-1.6493233426371i\)

gibsonj338  Nov 26, 2015
 #1
avatar
+5

z = (4 + 32i)^(1/5)  

Divide: 1 / 5 =0.2

Power: (4+32i) ^ 0.2 = 1.9198686+0.5714255i


Algebraic form:
z = 1.9198686+0.5714255i

Exponential form:
z = 2.0031032 × ei 16°34'30″

Trigonometric form:
z = 2.0031032 × (cos 16°34'30″ + i sin 16°34'30″)

Polar form:
r = |z| = 2.0031
φ = arg z = 16.575° = 16°34'30″ = 0.09208π

Guest Nov 24, 2015
 #2
avatar+1860 
+10
Best Answer

\((4+32i)^(1/5)\)

 

\(r=\sqrt(a^2+b^2)\)

 

\(r=\sqrt(4^2+32^2)\)

 

\(r=\sqrt(16+1024)\)

 

\(r=\sqrt1040\)

 

\(r=4\sqrt65\)

 

\(tan(\Theta)=b/a\)

 

\(tan(\Theta)=32/4\)

 

\(tan(\Theta)=8\)

 

\(\Theta=tan^-1(8)\)

 

\(\Theta ≈1.4464413322481\)

 

 

\(z=r*e^(i*\Theta)\)

 

\(z≈4\sqrt65*e^(i*1.4464413322481)\)

 

\(z^(1/5)≈(4\sqrt65*e^(i*1.4464413322481))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*1.4464413322481*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.0766143512748)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.0766143512748)+i*sin(1.0766143512748))\)

 

\(z≈2.003103242348*(0.4743116564868+i*0.88035700289064)\)

 

\(z≈0.9500952169545+i*1.763445966914\)

 

\(z≈0.9500952169545+1.763445966914i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*7.7296266394277))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*7.7296266394277*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.5459253278855)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.5459253278855)+i*sin(1.5459253278855))\)

 

\(z≈2.003103242348*(0.024868434927217+i*0.99969073264899)\)

 

\(z≈-0.049814942634829+i*0.0001545372385216\)

 

\(z≈-0.049814942634829+0.0001545372385216i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*14.012811946607))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*14.012811946607*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*1.6955283616309)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(1.6955283616309)+i*sin(1.6955283616309))\)

 

\(z≈2.003103242348*(-0.12440885450163+i*0.99223104009177)\)

 

\(z≈-0.24920377983349+i*1.9875412136019\)

 

\(z≈-0.24920377983349+1.9875412136019i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*20.2959972538))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*20.2959972538*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*4.059994507574)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(4.05994507574)+i*sin(4.05994507574))\)

 

\(z≈2.003103242348*(-0.60772245598411+i*-0.7941494925344)\)

 

\(z≈-1.2173308220295+i*1.5907634234047\)

 

\(z≈-1.2173308220295+1.5907634234047i\)

 

 

\(z^(1/5)≈(4\sqrt65*e^(i*26.5918256098))^(1/5)\)

 

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*26.5918256098*(1/5))\)

 

\(z^(1/5)≈2.003103242348*e^(i*5.315836512196)\)

 

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

 

\(z≈2.003103242348*(cos(5.315836512196)+i*sin(5.315836512196))\)

 

\(z≈2.003103242348*(0.56748448302716+i*-0.82338409112843)\)

 

\(z≈-1.1367300079339+i*-1.6493233426371\)

 

\(z≈-1.1367300079339+-1.6493233426371i\)

 

\(z≈-1.1367300079339-1.6493233426371i\)

gibsonj338  Nov 26, 2015

14 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.