Difference between revisions of "Manuals/calci/IMDIV"

Revision as of 03:46, 25 November 2013

IMDIV(z1,z2)

This function gives the division of two complex numbers.
This function used to remove the I (imaginary unit) from the denominator.
In IMDIV(z1,z2), where z1,z2 are the two complex numbers is in the form of z1=a+ib andz2=c+id, where a,b,c &d are real numbers i is the imaginary unit, i=sqrt(-1).
To do the division of complex number we have follow the steps:step1: 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. IMDIV(z1,z2)=(a+ib)/(c+id)=((a+ib)/(c+id))*((c-id)/(c-id))=[(ac+bd)/(c^2+d^2)]+[(bc-ad)i/[(c^2+d^2)]

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

@@ Line 8: / Line 8: @@
 *step 2: To find the conjugate of the denominator.
 *step 3:To mutiply the numerator and denominator with conjugate.
-i.e. IMDIV(z1,z2)=(a+ib)/(c+id)=((a+ib)/(c+id))*((c-id)/(c-id))
+i.e. IMDIV(z1,z2)=(a+ib)/(c+id)=((a+ib)/(c+id))*((c-id)/(c-id))=[(ac+bd)/(c^2+d^2)]+[(bc-ad)i/[(c^2+d^2)]
-                             =[(ac+bd)/(c^2+d^2)]+[(bc-ad)i/[(c^2+d^2)]
 ==Examples==
-#IMEXP("2+3i")=-7.315110094901102+1.0427436562359i
+#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
-#IMEXP("4-5i")=15.4874305606508+52.355491418482i
+#IMDIV("3-5i,2-6i")=0.9+0.2i
-#IMEXP("6")=403.428793492735
+#IMDIV("5","2+3i")=0.769-1.153i
-#IMEXP("2i")=-0.416146836547142+0.909297426825682i
+#IMDIV("1+i","2")=0.5+0.5i
-#IMEXP("0")=1 andIMEXP("0i")=1
 ==See Also==
@@ Line 22: / Line 20: @@
 *[[Manuals/calci/IMAGINARY  | IMAGINARY ]]
 *[[Manuals/calci/IMREAL  | IMREAL ]]
-*[[Manuals/calci/EXP  | EXP ]]
 ==References==
 [http://en.wikipedia.org/wiki/Exponential_function| Exponential function]