Difference between revisions of "Manuals/calci/IMDIV"
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<div style="font-size:30px">'''IMDIV(z1,z2)'''</div><br/> | <div style="font-size:30px">'''IMDIV(z1,z2)'''</div><br/> | ||
− | * | + | *<math>z1<math> and <math>z2<math> are complex numbers. |
==Description== | ==Description== | ||
*This function gives the division of two complex numbers. | *This function gives the division of two complex numbers. | ||
− | *This function used to remove the | + | *This function used to remove the <math>i<math> (imaginary unit) from the denominator. |
− | * | + | *<math>z1,z2</math> are the two complex numbers in the form of <math>z1=a+ib</math> and <math>z2=c+id</math>, where <math>a,b,c<math> & <math>d<math> are real numbers <math>i</math> is the imaginary unit, <math>i=\sqrt{-1}<math>. |
− | *To do the division of complex number we have follow the steps: | + | *To do the division of complex number we have follow the steps: |
− | + | step 1: We have to write the complex number is in the fraction form. | |
− | + | step 2: To find the conjugate of the denominator. | |
− | i.e. IMDIV(z1,z2)= | + | step 3: To mutiply the numerator and denominator with conjugate. |
+ | i.e. <math>IMDIV(z1,z2) = \frac{a+ib}{c+id} = \frac{a+ib}{c+id}*\frac{c-id}{c-id} =\frac{ac+bd}{c^2+d^2}+\frac{(bc-ad)i}{(c^2+d^2)} | ||
==Examples== | ==Examples== |
Revision as of 05:28, 25 November 2013
IMDIV(z1,z2)
- are the two complex numbers in the form of and , where Failed to parse (syntax error): {\displaystyle a,b,c<math> & <math>d<math> are real numbers <math>i} is the imaginary unit, <math>i=\sqrt{-1}<math>.
- To do the division of complex number we have follow the steps:
step 1: We have to write the complex number is in the fraction form. step 2: To find the conjugate of the denominator. step 3: To mutiply the numerator and denominator with conjugate.
i.e. <math>IMDIV(z1,z2) = \frac{a+ib}{c+id} = \frac{a+ib}{c+id}*\frac{c-id}{c-id} =\frac{ac+bd}{c^2+d^2}+\frac{(bc-ad)i}{(c^2+d^2)}
Examples
- IMDIV("4+2i","3-i")=(4+2i/3-i)*(3+i/3+i)=(12+10i+2i^2)/(3^2-i^2)=10+10i/10 (because i^2=-1)= 1+i/1=1+i
- IMDIV("3-5i,2-6i")=0.9+0.2i
- IMDIV("5","2+3i")=0.769-1.153i
- IMDIV("1+i","2")=0.5+0.5i
See Also