Difference between revisions of "Manuals/calci/IMSQRT"

Revision as of 22:18, 18 December 2013

IMSQRT(z)

This function gives square root of a complex number.
IMSQRT(z), Where z is the complex number is in the form of "x+iy".
where x&y are the real numbers.'i' imaginary unit . $i={\sqrt {-1}}$ .
The square root of a complex number is defined by Failed to parse (syntax error): {\displaystyle \sqrt{z}=\sqrt{x+iy}=\sqrt{r.e^iθ}=\sqrt{r}(cos(θ/2)+isin(θ/2)}
where r is the modulus of z. $r={\sqrt {x^{2}+y^{2}}}$
And θ is the argument of z.Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle θ=tan^-1(y/x)} also θ∈(-Pi(),Pi()].
We can use COMPLEX function to convert real and imaginary number in to a complex number.

Remarks

Examples

''''''' ''''

Description

This function calculates the square root of a complex number in a + bi or a + bj text format.

IMSQRT

The square root of a complex number is:

IMSQRT(IN)

where IN is the complex number

Let's see an example

I.e =IMSQRT(“2+i”) is 1.4553+0.34356i

@@ Line 8: / Line 8: @@
 *IMSQRT(z), Where z  is  the complex number is in the form of "x+iy".
 *where x&y are the real numbers.'i' imaginary unit .<math>i=\sqrt{-1}</math>.
-*The square root of a complex number is defined by  <math>\sqrt{z}=\sqrt{x+iy}=\sqrt{r.e^iθ}=\sqrt(r)(cos(θ/2)+isin(θ/2)</math>,where r is the modulus of z.<math>r=\sqrt{x^2+y^2}</math>
+*The square root of a complex number is defined by  <math>\sqrt{z}=\sqrt{x+iy}=\sqrt{r.e^iθ}=\sqrt{r}(cos(θ/2)+isin(θ/2)</math>
+*where r is the modulus of z.<math>r=\sqrt{x^2+y^2}</math>
 *And θ is the argument of z.<math> θ=tan^-1(y/x)</math> also θ∈(-Pi(),Pi()].
 *We can use COMPLEX function to convert real and imaginary number in to a complex number.