is calculated using Smith's algorithm with scaling, from Li et al. (2002), which ensures that no unnecessary overflow or underflow occurs at intermediate stages of the computation.

References

Li X S, Demmel J W, Bailey D H, Henry G, Hida Y, Iskandar J, Kahan W, Kapur A, Martin M C, Tung T and Yoo D J (2002) Design, implementation and testing of extended and mixed precision BLAS ACM Trans. Math. Soft. 28(2) 152–205

Parameters

Compulsory Input Parameters

1: $xr$ – double scalar
2: $xi$ – double scalar: $x_{r}$ and $x_{i}$ , the real and imaginary parts of $x$ , respectively.
3: $yr$ – double scalar
4: $yi$ – double scalar: $y_{r}$ and $y_{i}$ , the real and imaginary parts of $y$ , respectively.

Optional Input Parameters

None.

Output Parameters

1: $zr$ – double scalar
2: $zi$ – double scalar: $z_{r}$ and $z_{i}$ , the real and imaginary parts of $z$ , respectively.

Error Indicators and Warnings

None.

Accuracy

The result should be correct to machine precision.

Further Comments

The time taken by nag_complex_divide (a02ac) is negligible.

This function must not be called with

yr = 0.0

and

yi = 0.0

Example

This example finds the value of

(- 1.7 + 2.6 i) / (- 3.1 - 0.9 i)

Open in the MATLAB editor: a02ac_example

function a02ac_example


fprintf('a02ac example results\n\n');

xr = -1.7;
xi = 2.6;
yr = -3.1;
yi = -0.9;
x = xr+i*xi;
y = yr+i*yi;

[zr, zi] = a02ac(xr, xi, yr, yi);

z = zr+i*zi;

fprintf('The value of x/y, where\n');
fprintf('    x = ');
disp(x);
fprintf('and y = ');
disp(y);
fprintf('     is ');
disp(z);

a02ac example results

The value of x/y, where
    x =   -1.7000 + 2.6000i

and y =   -3.1000 - 0.9000i

     is    0.2812 - 0.9203i

PDF version (NAG web site, 64-bit version, 64-bit version)

Chapter Contents

Chapter Introduction

NAG Toolbox